Category: Inter 1st Year

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b)

I.

Question 1.
Find the slope of the tangent to the curve
y = 3x⁴ – 4x at x = 4.
Solution:
Equation of the curve is y = 3x⁴ – 4x
\(\frac{dy}{dx}\) = 12x³ – 4 dx
At x = 4, slope of the tangent = 12 (4)³ – 4
= 12 × 64 – 4
= 768 – 4
= 764

Question 2.
Find the slope of the tangent to the curve
y = \(\frac{x-1}{x-2}\) x ≠ 2 at x = 10.
Solution:
Equation of the curve is
y = \(\frac{x-1}{x-2}\)
= \(\frac{x-2+1}{x-2}\)
= 1 + \(\frac{1}{x-2}\)
\(\frac{dy}{dx}\) = 0 + \(\frac{(-1)}{(x-2)^{2}}=\frac{1}{(x-2)^{2}}\)
At x = 10, slope of the tangent = \(\frac{1}{(10-2)^{2}}\)
= –\(\frac{1}{64}\)

Question 3.
Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x co-ordinate is 2.
Solution:
Equation of the curve is y = x³ – x + 1
\(\frac{dy}{dx}\) = 3x² – 1
x = 2
Slope of the tangent at (x – 2) is
3(2)² – 1 = 3 x 4 – 1
= 12 – 1 = 11

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x co-ordinate is 3.
Solution:
Equation of the curve is y = x³ – 3x + 2
\(\frac{dy}{dx}\) = 3x² – 3
At x = 3, slope of the tangent = 3(3)² – 3
= 27 – 3 = 24

Question 5.
Find the slope of the normal to the curve
x = a cos³ θ, y = a sin³ θ at θ = \(\frac{\pi}{4}\).
Solution:
x = a cos³ θ
\(\frac{d x}{d \theta}\) = a(3 cos² θ) (-sin θ)
= -3a cos² θ. sin θ
y = sin³ θ
\(\frac{d y}{d \theta}\) = a (3 sin² θ) cos θ
= 3a sin² θ cos θ

At θ = \(\frac{\pi}{4}\), slope of the tangent = tan \(\frac{\pi}{4}\) = -1
Slope of the normal = – \(\frac{1}{m}\) = 1.

Question 6.
Find the slope of the normal to the curve
x = 1 – a sin θ, y = b cos θ at θ = \(\frac{\pi}{2}\).
Solution:
x = 1 – a sin θ
\(\frac{d x}{d \theta}\) = – a cos θ
y = b cos² θ dy
\(\frac{d y}{d \theta}\) = b(2 cos θ) (- sin θ) = -2b cos θ sin θ

\(\frac{2b}{a}\).sin θ
Slope of the normal = \(\frac{1}{m}=\frac{a}{2b \sin \theta}\)
At θ = \(\frac{\pi}{2}\), slope of the normal = \(\frac{-a}{2 b \sin \frac{\pi}{2}}\)
= \(\frac{-a}{2b.1}\)
= \(\frac{-a}{2b}\)

Question 7.
Find the points at which the tangent to the curve y = x3 – 3×2 – 9x + 7 is parallel to the x-axis.
Solution:
Equation of the curve is y = x³ – 3x² – 9x + 7
\(\frac{dy}{dx}\) = 3x² – 6x – 9 dx
The tangent is parallel to x-axis.
Slope of the tangent = 0
3x² – 6x – 9 = 0
x² – 2x – 3 = 0
(x – 3) (x + 1) = 0
x = 3 or -1
y = x³ – 3x² – 9x + 7
x = 3 ⇒ y = 27 – 27 – 27 + 7 = -20
x = -1, y = -1 – 3 + 9 + 7 = 12
The points required are (3, -20), (-1, 12).

Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution:
Equation of the curve is y = (x – 2)²
\(\frac{dy}{dx}\) = 2(x – 2)
Slope of the chord joining A(2, 0) and B(4, 4)
= \(\frac{4-0}{4-2}=\frac{4}{2}\) = 2.
The tangent is parallel to the chord.
2(x – 2) = 2
x – 2 = 1
x = 3
y = (x – 2)² = (3 – 2)² = 1
The required point is P(3, 1).

Question 9.
Find the point on the curve
y = x³ – 11x + 5 at which the tangent is y = x – 11.
Solution:
Equation of the curve is y = x³ – 11x + 5
\(\frac{dy}{dx}\) = 3x² – 11
The tangent is y = x – 11
Slope of the tangent = 3x² – 11 = 1
3x² = 12
x² = 4
x = ±2

y = x – 11
x = 2 ⇒ y = 2 – 11 = -9
The points on the curve is P(2, -9).

Question 10.
Find the equations of all lines having slope 0 which are tangents to the curve y = \(\frac{1}{x^{2}-2x+3}\).
Solution:
Equation of the curve is y = \(\frac{1}{x^{2}-2x+3}\)

Given slope of the tangent = 0

Equation the point is P(1, \(\frac{1}{2}\))
Slope of the tangent = 0
Equation of the required tangent is
y – \(\frac{1}{2}\) = 0(x – 1)
⇒ 2y – 1 = 0

II.

Question 1.
Find the equations of tangent and normal to the following curves at the points indicated against.
i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5).
Solution:
\(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10
At x = 0,
Slope of the tangent = 0 – 0 + 0 -10 = -10
Equation of the tangent is y – 5 = -10(x – 0)
= -10x
10x + y – 5 = 0
Slope of the normal = – \(\frac{1}{m}=\frac{1}{10}\)
Equation of the normal is y – 5 = \(\frac{1}{10}\) (x – 0)
10y – 50 = x ⇒ x – 10y + 50 = 0

ii) y = x³ at (1, 1).
Solution:
\(\frac{dy}{dx}\) = 3x²
At (1, 1), slope of the tangent = 3 (1)² = 3
Equation of the tangent at P(1, 1) is
y – 1 = 3(x – 1)
= 3x – 3
3x – y – 2 = 0
Slope of the normal = – \(\frac{1}{m}=-\frac{1}{3}\)
Equation of the normal is y – 1 = \(-\frac{1}{3}\)(x – 1)
3y – 3 = -x + 1
x + 3y – 4 = 0

iii) y = x² at (0, 0).
Solution:
Equation of the curve is y = x²
\(\frac{dy}{dx}\) = 2x
At P(0, 0), slope of the tangent = 2.0 = 0
Equation of the tangent is y – 0 = 0 (x – 0)
⇒ y = 0
The normal is perpendicular to the tangent.
Equation of the normal is x = k.
The normal passes through (0, 0) ⇒ k = 0
Equation of the normal is x = 0.

iv) x = cos t, y = sin t at t = \(\frac{\pi}{4}\).
Solution:
\(\frac{dx}{dt}\) = -sin t, \(\frac{dy}{dt}\) = cos t

Equation of the tangent is

Slope of the normal = –\(\frac{1}{m}=\frac{-1}{-1}\) = 1
Equation of the normal is y \(\frac{1}{\sqrt{2}}\) = x – \(\frac{1}{\sqrt{2}}\)
i.e., x – y = 0

v) y = x² – 4x + 2 at (4, 2).
Solution:
Equation of the curve is y = x² – 4x + 2
\(\frac{dy}{dx}\) = 2x – 4
At P(4, 2), slope of the tangent =2.4 – 4
= 8 – 4 = 4
Equation of the tangent at P is
y – 2 = 4(x – 4)
= 4x – 16
4x – y – 14 = 0
Slope of the normal = –\(\frac{1}{m}=-\frac{1}{4}\)
Equation of the normal at P is
y – 2 = \(-\frac{1}{4}\) (x – 4)
⇒ 4y – 8 = -x + 4
⇒ x + 4y – 12 = 0

vi) y = \(-\frac{1}{1+x^{2}}\) at (0, 1)
Solution:
Equation of the curve is y = \(-\frac{1}{1+x^{2}}\)
\(\frac{dy}{dx}\) = \(-\frac{1}{(1+x^{2})^{2}}\)
At (0, 1), x = 0, slope of the tangent = 0
Equation of the tangent at P(0, 1) is
y – 1 = 0(x – 0)
y = 1
The normal is perpendicular to the tangent.
Equation of the normal can be taken at x = 10.
The normal passes through P(0, 1) ⇒ 0 = k
Equation of the normal at P is x = 0.

Question 2.
Find the equations of tangent and normal to the curve xy = 10 at (2, 5).
Solution:
Equation of the curve is xy = 10.
y = \(\frac{10}{x}\); \(\frac{dy}{dx}=\frac{10}{x^{2}}\)
At P(2, 5), f'(x₁) = –\(\frac{10}{4}=-\frac{5}{2}\)
Equation of the tangent is
y – y₁ = f'(x₁) (x – x₁)
y – 5 = – \(\frac{5}{2}\) (x – 2)
2y – 10 = -5x + 10
5x + 2y – 20 = 0
Equation of the normal is
y – y₁ = \(\frac{1}{f'(x_{1})}\)(x – x₁)
y – 5 = \(\frac{5}{2}\) (x – 2)
5y – 25 = 2x – 4
i.e., 2x – 5y + 21 = 0.

Question 3.
Find the equations of tangent and normal to the curve y = x³ + 4x² at (-1, 3).
Solution:
Equation of the curve is y = x³ + 4x²
\(\frac{dy}{dx}\) = 3x² + 8x
At P(-1, 3),
Slope of the tangent
= 3(-1)² + 8(-1)
= 3 – 8 = -5

Equation of the tangent at P(-1, 3) is
y – y₁ = f'(x₁) (x – x₁)
y – 3 = -5(x + 1) = -5x – 5
5x + y + 2 = 0
Equation of the nonnal at P is
y – y₁ = –\(\frac{1}{f'(x_{1})}\) (x – x₁)
y – 3 = \(\frac{1}{5}\) (x + 1)
5y – 15 = x + 1
x – 5y + 16 = 0

Question 4.
If the slope of the tangent to the curve x² – 2xy + 4y = 0 at a point on it is –\(\frac{3}{2}\), then find the equations of tangent and normal at that point.
Solution:
Equation of the curve is
x² – 2xy + 4y = 0 ………… (1)
Differentiating w.r.to x

2x – 2y = -3x + 6; 5x – 2y = 6
2y = 5x – 6 ……. (2)
P(x, y) is a point on (1)
x² – x(5x – 6) + 2(5x – 6) = 0
x² – 5x² + 6x + 10x – 12 = 0
-4x² + 16x – 12 = 0
-4(x² – 4x + 3) = 0
x² + 4x + 3 = 0
(x – 1) (x – 3) = 0
x – 1 = 0 or x – 3 = 0
∴ x = 1 or x = 3

Case (i): x = 1
Substituting in (1)
1 – 2y + 4y = 0
2y = -1 ⇒ y = –\(\frac{1}{2}\)
The required point is P(1, –\(\frac{1}{2}\))
Equation of the tangent is
y + \(\frac{1}{2}\) = –\(\frac{3}{2}\)(x – 1)
\(\frac{2y+1}{2}=\frac{-3(x-1)}{2}\)
2y + 1 = -3x + 3
3x + 2y – 2 = 0
Equation of the normal isy + \(\frac{1}{2}=\frac{2}{3}\)(x – 1)
\(\frac{2y+1}{2}=\frac{2}{3}\) (x – 1)
6y + 3 = 4x – 4
4x – 6y – 7 = 0

Case (ii) : x = 3
Substituting in (1), 9 – 6y + 4y = 0
2y = 9 ⇒ y = \(\frac{9}{2}\)
∴ The required point is (3, \(\frac{9}{2}\))
Equation of the tangent is
y – \(\frac{9}{2}=-\frac{3}{2}\) (x – 3)
\(\frac{2y-9}{2}=\frac{-3(x-3)}{2}\)
2y – 9 = -3x + 9
3x + 2y- 18 = 0
Equation of the normal is y – \(\frac{9}{2}=\frac{2}{3}\) (x – 3)
\(\frac{2y-9}{2}=\frac{2(x-3)}{3}\)
6y – 27 = 4x – 12
i.e., 4x – 6y + 15 = 0.

Question 5.
If the slope of the tangent to the curve y = x log x at a point on it is \(\frac{3}{2}\), then find the equations of tangent and normal at that point.
Solution:
Equation of the curve is y = x log x
\(\frac{dy}{dx}\) = x. \(\frac{1}{x}\) + log x.1 = 1 + log x.
Given 1 + log x = \(\frac{3}{2}\)
log_e x =\(\frac{1}{2}\) ⇒ x = e^½ = √e
y = √e . log .√e = \(\frac{\sqrt{e}}{2}\)
The required point is P (√e, \(\frac{\sqrt{e}}{2}\))
Equation of the tangent is y \(\frac{\sqrt{e}}{2}=\frac{3}{2}\)(x – √e)
\(\frac{2 y-\sqrt{e}}{2}=\frac{3(x-\sqrt{e})}{2}\)
2y – √e = 3x – 3 √e
3x – 2y – 2√e = 0
Equation of the normal is
y – y₁ = – \(\frac{1}{f'(x_{1})}\)(x – x₁)
y – \(\frac{\sqrt{e}}{2}=-\frac{2}{3}\)(x – √e)
\(\frac{2 y-\sqrt{e}}{2}=-\frac{2}{3}\)(x – √e)
6y – 3√e = -4x + 4√e
i.e., 4x + 6y – 7√e =0

Question 6.
Find the tangent and normal to the curve y = 2e^-x/3 at the point where the curve meets the Y-axis.
Solution:
Equation of the curve is y = 2e^-x/3
Equation of Y-axis is x = 0
y = 2.e° = 2.1 = 2
Required point is P(0, 2)
\(\frac{dy}{dx}\) = 2(-\(\frac{1}{3}\)) . e^-x/3
When x = 0, slope of the tangent = –\(\frac{2}{3}\) .e° = \(\frac{-2}{3}\)
Equation of the tangent at P is
y – y₁ = f'(x₁) (x – x₁)
y – 2 = –\(\frac{2}{3}\) (x – 0)
3y – 6 = -2x
2x + 3y – 6 = 0
Equation of the normal is
y – y₁ = –\(\frac{1}{f'(x_{1})}\) (x – x₁
y – 2 = \(\frac{3}{2}\) (x – 0)
2y – 4 = 3x; 3x – 2y + 4 = 0

III.

Question 1.
Show that the tangent at P(x₁, y₁) on the curve √x + √y = √a is yy₁^-½ + xx₁^-½ = a^½.
Solution:
Equation of the curve is √x + √y = √a
Differentiating w.r.to x

Slope of the tangent at P(x₁ y₁) = –\(\frac{\left(y_{1}\right)^{1 / 2}}{\left(x_{1}\right)^{1 / 2}}\)
Equation of the tangent at P is

= x₁^½ + y₁^½
x. x₁^-½ + y. y₁^-½ = a^½
(P is a point on the curve)
Equation of the tangent at P is
y. y₁^-½ + x. x₁^-½ = a^½

Question 2.
At what points on the curve x² – y² = 2, the slopes of the tangents are equal to 2?
Solution:
Equation of the curve is x² – y² = 2 ………. (1)
Differentiating w.r.to x
2x – 2y.\(\frac{dy}{dx}\) = 0 ⇒ \(\frac{dy}{dx}=\frac{x}{y}\)
Slope of the tangent = \(\frac{dy}{dx}\) = 2
∴ \(\frac{x}{y}\) = 2 ⇒ x = 2y
Substituting in (1), 4y² – y² = 2
3y² = 2
y ² = \(\frac{2}{3}\) ⇒ y = ± \(\sqrt{\frac{2}{3}}\)
x = 2y = ± 2 \(\sqrt{\frac{2}{3}}\)
∴ The required points are

Question 3.
Show that the curves x² + y² = 2 and 3x² + y² = 4x have a common tangent at the point (1, 1).
Solution:
Equation of the first curve is x² + y² = 2
Differentiating w.r.to x

At P (1, 1) slope of the tangent = \(\frac{-1}{1}\) = -1
Equation of the second curve is 3x² + y² = 4x.
Differentiating w.r.to x, 6x + 2y.\(\frac{dy}{dx}\) = 4
2y.\(\frac{dy}{dx}\) = 4 – 6x
\(\frac{dy}{dx}=\frac{4-6x}{2y}=\frac{2-3x}{y}\)
At P( 1, 1) slope of the tangent = \(\frac{2-3}{1}\) = –\(\frac{1}{1}\) = -1
The slope of the tangents to both the curves at P( 1, 1) are same and pass through the same point (1, 1)
∴ The given curves have a common tangent at P (1, 1)

Question 4.
At a point (x₁, y₁) on the curve x³ + y³ = 3axy, show that the tang;ent is
(x₁² – ay₁) x+ (y₁² – ax₁)y = ax₁y₁
Solution:
Equation of the curve is x³ + y³ = 3axy
Differentiating w. r. to x

Slope of the tangent P(x₁, y₁) = –\(\frac{\left(x_{1}^{2}-a y_{1}\right)}{\left(y_{1}^{2}-a x_{1}\right)}\)
Equation of the tangent at P(x₁, y₁) is
y(y – y₁) = –\(\frac{\left(x_{1}^{2}-a y_{1}\right)}{\left(y_{1}^{2}-a x_{1}\right)}\)(x – x₁)
y(y₁² – ax₁) – y₁(y₁² – ax₁) = – x(x₁² – ay₁) + x₁(x₁² – ay₁)
x₁(x₁² – ay₁) + y₁(y₁² – ax₁)
= x₁(x₁² – ay₁) + y₁(y₁² – ax₁)
= x₁³ – ax₁y₁ + y₁³ – ax₁y₁
= x₁³ + y₁³ – 2ax₁y₁
3ax₁y₁ – 2ax₁y₁ (P is a point on the curve)
= ax₁y₁

Question 5.
Show that the tangent at the point P (2, -2) on the curve y (1 – x) = x makes intercepts of equal length on the co-ordinate axes and the normal at P passes through the origin.
Solution:
Equation of the curve is
y (1 – x) = x
y = \(\frac{x}{1-x}\)
Differentiating w.r. to x

Equation of the tangent at P is
y + 2 = +(x – 2) = x – 2; x – y = 4
\(\frac{x}{4}-\frac{y}{4}\) ⇒ \(\frac{x}{4}-\frac{y}{(-4)}\) = 1
∴ a = 4, b = – 4
∴ The tangent makes equal intercepts on the co-ordinate axes but they are in opposite in sign. Equation of the normal at P is
y – y₁ = \(\frac{1}{f'(x_{1})}\) (x – x₁)
y + 2 = -(x – 2)= -x + 2
x + y = 0
There is no constant term in the equation.
∴ The normal at P(2, -2) passes through the origin.

Question 6.
If the tangent at any point on the curve x^2/3 + y^2/3 = a^2/3 intersects the coordinate axes in A and B then show that length AB is a constant.

Solution:
Equation of the curve is
x^2/3 + y^2/3 = a^2/3
Differentiating w.r. to x

Equation of the tangent at P (x₁, y₁) is

AB = a = constant.

Question 7.
If the tangent at any point P on the curve x^m yⁿ = a^m+n (mn ≠ 0) meets the co-ordinate axes in A, B, then show that AP: PB is a constant.

Solution:
Equation of the curve is x^m.yⁿ = a^m+n
Differentiating w.r. to x

Slope of the tangent at P(x₁, y₁) = –\(\frac{my_{1}}{nx_{1}}\)
Equation of the tangent at P is

Co-ordinates of A are [\(\frac{m+n}{m}\).x₁, o] and B are [0, \(\frac{m+n}{m}\).y₁]
Let P divide AB in the ratio k : l
Co-ordinates of P are


Dividing (1) by (2) \(\frac{l}{k}=\frac{m}{n}\) ⇒ \(\frac{k}{l}=\frac{n}{m}\)
∴ P divides AB in the ratio n : m
i.e., AP : PB = n : m = constant.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(a)

I.

Question 1.
Find ∆y and dy for the following functions for the values of x and ∆x which are shown against each of the functions,
i) y = x² + 3x + 6, x = 10, ∆x = 0.01.
Solution:
∆y = f(x + ∆x) – f(x)
= f(10.01)-f(10)
= E(10.01)² + 3(10.01) + 6] – [10² + 3(10) + 6]
= 100.2001 +30.03 + 6 – 100 – 30 – 6
= 0.2001 + 0.03
= 0.2301
y = x² + 3x + 6
dy = (2x + 3) dx
= (2.10 + 3) (0.01) = 0.23

ii) y = e^x + x, x = 5 and ∆x = 0.02
Solution:
∆y = f(x + ∆x) – f(x)
= f(5 + 0.02) – f(5)
= f(5.02) – f(5)
= (e^5.02 + 5.02) – (e⁵ + 5)
= e^5.02 – e⁵ + 0.02
= e⁵ (e^0.02 – 1) + 0.02
dy = f'(x) ∆x = (e^x + 1) ∆x
= (e⁵ + 1) (0.02)

iii) y = 5x² + 6x + 6, x = 2 and ∆x = 0.001
Solution:
∆y’= f(x + ∆x) – f(x)
= f(2 +0.001) – f(2)
= f(2.001) – f(2)
= (5(2.001)² + 6(2.001) + 6) – (5(2)² + 6(2) +6)
= 20.0200 + 12.0060 + 6 – 20 – 12 – 6
= 0.026005
dy = f'(x) ∆x = (10x + 6) ∆x
= (26) (0.001) = 0.0260.

iv) y = 2 \(\frac{1}{x+2}\) x = 8 and ∆x = 0.02
Solution:
f(x) = \(\frac{1}{x+2}=\frac{1}{10}\) = 0.1000
f(x + ∆x) = \(\frac{1}{x+\Delta x+2}=\frac{1}{10+0.02}=\frac{1}{10.02}\) = 0.0998
∆y = f(x + ∆x) – f(x)
= \(\frac{1}{x+\Delta x+2}-\frac{1}{1+x}=\frac{1}{10.02}=\frac{1}{10}\)
= 0.0998 003992 – 0.1000 = – 0.0001996
dy = f'(x) ∆x = \(\frac{-1}{1+x^{2}}\) ∆x
= \(\frac{-1}{100}\)(0.02) = -0.0002

v) y = cos (x), x = 60° and ∆x = 1°
Solution:
∆y = f(x + ∆x) – f(x)
= cos (x + ∆x) – cos x
= cos (60° + 1°) – cos 60°
= cos 61° – cos 60°
= 0.4848 – \(\frac{1}{2}\) = 0.4848 – 0.5 = – 0.0152
dy = f'(x) ∆x
= — sin x ∆x
= – sin 60°(1°) = \(\frac{-\sqrt{3}}{2}\) (0.0174)
= – (0.8660) (0.0174) = – 0.0151.

II.

Question 1.
Find the approximations of the following.
i) √82
Solution:
82 = 81 + 1 = 81(1 + \(\frac{1}{81}\))
∴ x = 81, ∆x = 1, f(x) = 77
dy = f'(x). ∆x = \(\frac{1}{2\sqrt{x}}\). ∆x = \(\frac{1}{2\sqrt{81}}\).1
= \(\frac{1}{18}\) = 0.0555
f(x + δx) – f(x) ≅ dy
f(x + δx) ≅ f(x) + dy
= √81 + 0.0555
= 9 + 0.0555
i.e., √82 = 9.0555 = 9.056

ii) \(\sqrt[3]{65}\)
Solution:
Let x = 64, ∆x = 1, f(x) = \(\sqrt[3]{x}\)
f'(x) = \(\frac{1}{3}\)x^-2/3
f(x + ∆x) ≅ f(x) + f'(x)∆x
\(\sqrt[3]{65}\) ≅ \(\sqrt[3]{x}\) + \(\frac{1}{3}\)x^-2/3 ∆x
≅ \(\sqrt[3]{65}+\frac{1}{3}\)(4)^-2/3(I)
≅ 4 + \(\frac{1}{3}\)(\(\frac{1}{16}\))
≅ 4 + \(\frac{1}{48}\)
≅ \(\frac{192+1}{48}\)
≅ \(\frac{193}{48}\) ≅ 4.0208

iii) \(\sqrt{25.001}\)
Solution:
Letx = 25, ∆x- 0.001
f(x) = √x
dy = f'(x) ∆x
= \(\frac{1}{2\sqrt{x}}\) ∆x = \(\frac{1}{2\sqrt{25}}\) (0.001) = \(\frac{0.001}{10}\) = 0.0001
f(x + ∆x) ≅ f(x) + dy
≅ √25 + 0.0001
≅ 5.0001

iv) \(\sqrt[3]{7.8}\)
Solution:
Let x = 8, ∆x = -0.2, f(x) = \(\sqrt[3]{x}\)
dy = f'(x). ∆x
= \(\frac{1}{3}\)x^-2/3. ∆x = \(\frac{1}{3x^{2/3}}\) . ∆x
dy = \(\frac{1}{3(8)^{2/3}}\)(-0.2)
= – \(\frac{1}{3}\)
\(=-\frac{0.2}{3 \times 4}=-\frac{0.2}{12}\)
f(x + δx) – (x) ≅ dy
f(x + δx) ≅ f(x) + dy
= \(\sqrt[3]{8}\) – 0.0166
= 2 – 0.0166
= 1.9834
∴ \(\sqrt[3]{7.8}\) = 1.9834

v) sin (62°)
Solution:
Let x = 60°, ∆x = 2°, f(x) = sin x
dy = f'(x) ∆x
= cosx ∆x
= cos 60° ∆x
= \(\frac{1}{2}\) (2°)
= \(\frac{1}{2}\) 2(0.0174) = 0.0174

f(x + ∆x) ≅ f(x) + dy
≅ sin 60° + 0.0174
≅ \(\frac{\sqrt{3}}{2}\) + 0.0174
≅ 0.8660 + 0.0174
≅ 0.8834

vi) cos (60° 5′)
Solution:
Let x = 60°, Ax = 5′ = \(\frac{5}{60}\)×\(\frac{\pi}{180}=\frac{\pi}{2160}\)
= 0.001453
f(x) = cos x
dy = f'(x) ∆x = – sin x ∆x
= – sin 60° (0.001453)
= \(\frac{-\sqrt{3}}{2}\) (0.001453)
= – 0.8660 (0.001453)
= -0.001258

f(x + ∆x) ≅ f(x) + dy
≅ cos x + dy
≅ cos 60° + 0.001258
≅ 0.5 – 0.001258
≅ 0.4987.

vii) \(\sqrt[4]{17}\)
Solution:
Let x – 16, ∆x = 1, f(x) = \(\sqrt[4]{x}\) = x^¼
dy = f'(x) ∆x
= \(\frac{1}{4}\) x^¼-1 ∆x
= \(\frac{1}{4}\) x^-3/4 ∆x
= \(\frac{1}{4}\) (16)^-3/4 (I)
= \(\frac{1}{32}\) = 0.0312

f(x + ∆x) ≅ f(x) + dy
≅ \(\sqrt[4]{x}\) + 0.0312
≅ 2 + 0.0312
≅ 2.0312

Question 2.
If the increase in the side of a square is 4% then find the approximate percentage of increase in the area of the square.
Solution:
Let x be the side and A be the area of square
A = x²

Question 3.
The radius of a sphere is measured as 14 cm. Later it was found that there is an error 0.02 cm in measuring the radius. Find the approximate error in surface of the sphere.
Solution:
Let s be the surface of the sphere
r’ = 14, ∆r = 0.02
s = 4πr²
∆s = 4π 2r ∆r
∆s = 8π (14) (0.02)
= 2.24π
= 2.24 (3.14)
= 7.0336.

Question 4.
The diameter of a sphere is measured to be 40 cm. If an error of 0.02 cm is made in it, then find approximate errors in volume and surface area of the sphere.
Solution:
Let v be the value of sphere
v = \(\frac{4}{3}\) πr³ = \(\frac{4 \pi}{3}\)[latex]\frac{d}{2}[/latex]³
= \(\frac{4 \pi}{3} \frac{d^{3}}{8}=\frac{\pi d^{3}}{6}\)
∆v = \(\frac{\pi}{6}\)3d² ∆d
= \(\frac{\pi}{2}\) (40)² (0.02)
= π(1600) (0.01)
= 16π.

Surface Area s = 4πr²
s = 4π [latex]\frac{d}{2}[/latex]²
s = 4π\(\frac{d^{2}}{4}\)
s = πd²
∆s = π2d ∆d
= π2d (40) (0.02)
= 1.6π.

Question 5.
The time t, of a complete oscillation of a simple pendulum of length l is given by t = \(2 \pi \sqrt{\frac{1}{g}}\) where g is gravitational constant. Find the approximate percen-tage of error in t when the percentage of error l is 1%.
Sol. Given t = \(2 \pi \sqrt{\frac{1}{g}}\)
log t = log 2π + \(\frac{1}{2}\) {(log l – log)}

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Applications of Derivatives Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Applications of Derivatives Important Questions

Question 1.
Find dy and ∆y of y = f(x) = x² + x at x = 10 when ∆x = 0.1.
Solution:
As change in y – f(x) is given by ∆y = f(x + ∆x) – f(x), this change at x = 10 with ∆x = 0.1 is
∆y = f(10.1) – f(10)
= {(10.1)² + 10.1} – {10² + 10}
= 2.11.
Since dy = f'(x) ∆x, dy at x = 10 with ∆x = 0.1 is dy = {(2)(10) + 1} 0.1 = 2.1
(since \(\frac{\mathrm{d} y}{\mathrm{dx}}\) = 2x + 1).

Question 2.
Find ∆y and dy for the function y = cos (x) at x = 60° with ∆x = 1° .
Solution:
For the given problem ∆y and dy at x = 60° with ∆x = 1° are
∆y = cos (60° + 1°) – cos (60°) …………. (1)
and dy = -sin(60°) (1°) ………….. (2)
Cos (60°)= 0.5,
Cos (61°)= 0.4848,
Sin (60°) = 0.8660,
1° = 0.0174 radians
∴ ∆y = -0.0152 and
dy = -0.0150.

Question 3.
The side of a square is increased from 3 cm to 3.01 cm. Find the approximate increase in the area of the square.
Solution:
Let x be the side of a square and A be its area.
Then A = x². …… (1)
Clearly A is a function of x. As the side is increased from 3 cm to 3.01 cm we can take x – 3 and ∆x = 0.01 to compute the approximate increase in the area of square. The approximate value of change in area is
∆A ≈ \(\frac{\mathrm{dA}}{\mathrm{dx}}\) ∆x
In view of equation (1), the equation (2) becomes
∆A ≈ 2x∆x
Hence the approximate increase in the area when the side is increased from 3 to 3.01 is
∆A ≈ 2(3)(0.01) = 0.06

Question 4.
If the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere.
Solution:
Let r be the radius of a sphere and V be its volume. Then
V = \(\frac{4 \pi \pi^{2}}{3}\) …………….. (1)
Here V is a function of r. As the radius is increased from 7 cm to 7.02, we can take r = 7 cm and ∆r = 0.02 cm. Now we have to find the approximate increase in the volume of the sphere.
∴ ∆V ≈ \(\frac{\mathrm{dV}}{\mathrm{dr}}\) ∆r = 4πr² ∆r.
Thus, the approximate increase in the volume of the sphere is \(\frac{4(22)(7)(7)(0.02)}{7}\) = 12.32 cm³.

Question 5.
If y = f(x) = k xⁿ then show that the approximate relative error (or increase) in y is n times the relative error (or increase) in x where n and k are constants.
Solution:
The approximate relative error (or increase) in y by the equation (2) of if a number A is very close to a number B but it is not equal to B then A is called an approximate value of B is (\(\frac{f^{\prime}(x)}{f(x)}\)) ∆x = \(\frac{k n x^{n-1}}{k x^{n}}\) ∆x = n(\(\frac{\Delta x}{x}\) = n)
Hence the approximate relative error in y = kxⁿ is n times the relative error in x.

Question 6.
If the increase in the side of a square is 2% then find the approximate percentage of increase in its area.
Solution:
Let x be the side of a square and A be its area.
Then A = x².
Approximate percentage error in area A
= (\(\frac{\frac{\mathrm{dA}}{\mathrm{dx}}}{\mathrm{A}}\)) × 100 × ∆x(by (3) of if a number
A is very close to a number B but it is not equal to B then A is called an approximate value of B with f = A)
= \(\frac{100(2 x) \Delta x}{x^{2}}\) = \(\frac{200 \Delta x}{x}\) = 2(2) = 4
(∵ \(\frac{\Delta x}{x}\) × 100 = 2

Question 7.
If an error of 0.01 cm is made in measuring the perimeter of a circle and the perimeter is measured as 44 cm then find the approximate error and relative error in its area.
Solution:
Let r, p and A be the radius, perimeter and area of the circle respectively. Given that p = 44 cm and ∆p = 0.01. We have to find approximation of ∆A and \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\). Note that A = πr² which is a function of r. As p and ∆p are given we have to transform A = πr² into the form A = f(p). This can be achieved by using the relation, perimeter 2πr = p.
∴ A = π(\(\frac{p}{2 \pi}\))² = \(\frac{p^{2}}{4 \pi}\)
Hence the approximate error in
A = \(\frac{d A}{d p}\)∆p = \(\frac{2 p}{4 \pi}\)∆p = \(\frac{P}{2 \pi}\)∆p
The approximate error in A when p = 44 and ∆p = 0.01 = \(\) (0.01) = 0.07
The approximate relative error

= 0.0004545.

Question 8.
Find the approximate value of \(\sqrt[3]{999}\)
Solution:
This problem can be answered by
f(x + ∆x) ≈ f(x) + f'(x) ∆x …………. (1)
with x = 1000 and ∆x = -1. The reason for taking x = 1000 is to make the calculation of f(x) simpler when f(x) = \(\sqrt[3]{x}\), Suppose
y = f(x) = \(\sqrt[3]{x}\)
The equation (1) becomes
f(x + ∆x) ≈ f(x) = \(\frac{1}{3 x^{\frac{2}{3}}}\) ∆x
Hence f(1000 – 1)
≈ f(1000) + \(\frac{1}{3(1000)^{2 / 3}}\) (-1) = 9.9967.

Question 9.
Find the slope of the tangent to the following curves at the points as indicated.
i) y = 5x² at (-1, 5)
ii) y = \(\frac{1}{x – 1}\) (x ≠ 1) at [3, \(\frac{1}{2}\)]
iii) x = a secθ, y = a tanθ at θ = \(\frac{\pi}{6}\)
iv) (\(\frac{x}{a}\))ⁿ + (\(\frac{x}{b}\))ⁿ = 2 at (a, b)
Solution:
i) y = 5x², then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 10x
Slope of the tangent at the given points
(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_(-1.5) -10(-1) = -10

ii) y = \(\frac{1}{x – 1}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{(x-1)^{2}}\)
Slope of the tangent at the given point is

Question 10.
Find the equations of the tangent and the normal to the curve y = 5x⁴ at the point (1, 5)
Solution:
y = 5x⁴ implies that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 20x³
Slope of the tangent to the curve at (1, 5) is
(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(1, 5)} = 20(1)⁵ = 20
Equation of the normal of (1, 5) is
y – 5 = 20 (x – 1) = 20x – 20
y = 20x – 15
Slope of the normal to the curve at (1, 5) is
–\(\frac{1}{m}\) = –\(\frac{1}{20}\)
Equation of the normal of (1, 5) is
y – 5 = –\(\frac{1}{20}\) (x – 1)
20 y – 100 = -x + 1
x + 20y = 101
(or) 20y = 101 – x

Question 11.
Find the equations of the tangent and the normal to the curve y⁴ = ax³. at (a, a).
Solution:
Given curve is y⁴ = ax³
Differentiating w.r.to. x
4y³ . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3ax²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{3 a x^{2}}{4 y^{3}}\)
Slope of the tangent at (a, a) = \(\frac{3 a \cdot a^{2}}{4 a^{3}}\) = \(\frac{3}{4}\)
Slope of the normal at (a, a) = –\(\frac{1}{m}\) = –\(\frac{4}{3}\)
Equation of the tangent at (a, a) is
y – a = \(\frac{3}{4}\)(x – a)
4y – 4a = 3x – 3a
4y = 3x + a
Equation of the normal at (a, a) is 4
y – a = – \(\frac{4}{3}\)(x – a)
3y – 3a = -4x + 4a
3y + 4x = 7a

Question 12.
Find the equations of the tangent to the curve y = 3x² – x³, where it meets the X-axis.
Solution:
Equation of the curve is y = 3x² – x³
Equation of X – axis is y = 0
For points is intersection of the curve and X-axis
3x² – x³ = 0 ⇒ x² (3 – x) = 0
x = 0, x = 3
The curve crosses X-axis 0(0, 0) and A(3, 0)
y = -3x² – x³
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6x – 3x²
At O(0, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(0, 0)} = 0
Equation of the tangent at (0, 0) is y – 0, 0(x – 0) i.e., y = 0
i.e., x-axis is the tangent to the curve at O(0, 0)
At A(3, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(3, 0)}
= 6.3 – 3.3²
= 18 – 27
= -9
Equation of the tangent at A(3, 0) is
y – 0 = -9(x – 3) = -9x + 27
(or) 9x + y = 27

Question 13.
Find the points at which the curve t y = sin x has horizontal tangents.
Solution:
y = sin x

A tangent is horizontal if and anal its slope is
cos x = 0 ⇒ x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
Hence the given curve has horizontal tangents at points (x₀, y₀)
⇔ x₀ = (2n + 1) . \(\frac{\pi}{2}\) and
y₀ = (-1)ⁿ for same n ∈ Z

Question 14.
Verify whether the curve y = f(x) = x^1/3 has a vertical tangent at the point with x = 0.
Solution:

The function has a verified tangent at the point whose x co-ordinate is 0.

Question 15.
Find whether the curve y = f(x)= x^2/3 has a vertical tangent at x = 0.
Solution:

Thus left handed be normal \(\frac{1}{h^{1 / 3}}\) as h → 0 is -∞
While the right handed limit is ∞.
Hence does not exist. The vertical tangent does not exist.
At the point x = 0.

Question 16.
Show that the tangent at any point 0 on the curve x = c sec θ, y = c tan θ is y sin θ = x – c cos θ.
Solution:
x = c sec θ, y = c tan θ

Question 17.
Show that the area of the triable formed by the tangent at any point on the curve xy = c (c ≠ 0) with the coordinate axis is constant.
Solution:
Observe that c ≠ 0
If c = 0 the equation xy = 0 represent the coordinate circle which is against the definite.
Let P (x₁, y₁) be a point on the curve xy = c
y = \(\frac{c}{x}\) = 1, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – \(\frac{c}{x^{2}}\)
Equation of the tangent at (x₁, y₁) is
y – y₁ = – \(\frac{c}{x^{2}}\) (x – x₁)
x²y – x₁² = -cx + cx₁
cx + x₁² . y = x₁² + cx₁
= cx₁ + cx₁ (x₁y₁ = c)
= 2cx₁
\(\frac{c x}{2 c x_{1}}\) + \(\frac{x_{1}^{2} y}{2 c x_{1}}\) = 1
\(\frac{x}{2 x_{1}}\) + \(\frac{y}{\left(\frac{2 c}{x_{1}}\right)}\) = 1
Area of the triangle formed with co-ordinate axes
= \(\frac{1}{2}\) |OA . OB|
= \(\frac{1}{2}\) (2x₁) (\(\frac{2 c}{x_{1}}\)) = 2c = constant

Question 18.
Show that the equation of the tangent to the curve (\(\frac{x}{a}\))ⁿ + (\(\frac{y}{b}\))ⁿ = 2 (a ≠ 0, b ≠ 0) at the point (a, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 2
Solution:

Equation of the tangent to the curve at the point (a, b) is
y – b = \(\frac{-b}{a}\) (x – a)
\(\frac{y}{b}\) – 1 = – \(\frac{x}{a}\) + 1
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 2

Question 19.
Show that the length of the sub normal at point on the curve y² = 4ax is a constant.
Solution:
Equation of the curve is y² = 4ax
Differentiating w.r.to x
2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4a
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{4a}{2y}\) = \(\frac{2a}{y}\)
Length of the sub-normal \(\left|\frac{y d y}{d x}\right|=\left|y \cdot \frac{2 a}{y}\right|\)
= 2a = constant

Question 20.
Show that the length of the Sub tangent at any point on the curve y = a^x (a > 0) is a constant.
Solution:
Equation of the curve is y = a^x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a^x log a = y. log a
Length of the sub-tangent
= \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\frac{1}{\log a}\) constant

Question 21.
Show that the square of the length of subtangent at any point on the curve by² = (x + a)³ (b ≠ 0) varies with the length of the subnormal at that point.
Solution:
Differentiating by² = (x + a)³
w.r.t x, we get
2by y’ = 3(x + a)²
∴ The length of the subnormal at any point (x, y) on the curve
= |y y’| = |\(\frac{3}{2 b}\)(x + a)²| ………………… (1)
The square of the length of subtangent

∴ (length of the subnormal)² ∝ (length of subnormal).

Question 22.
Find the value of k, so that the length of the subnormal at any point on the curve y = a^{1 – k}x^k is a constant.
Solution:
Differentiating y = a^{1 – k}x^k with respect to x,
we get y’ = ka^{1 – k}x^{k – 1}
Length of subnormal at any point P(x, y) on the curve
= |y y’| = |yka^{1 – k}x^{k – 1}|
= |k a^{1 – k}x^ka^{1 – k}x^{k – 1}|
= |ka^2-2kx^2k-1|
In order to make these values a constant, we should have 2k – 1 = 0 i.e., k = \(\frac{1}{2}\).

Question 23.
Find the angle between the curves xy = 2 and x² + 4y = 0.
Solution:
First we find the points of intersection of xy = 2 and x² + 4y = 0
y = \(\frac{-x^{2}}{4}\)
But xy = 2 ⇒ x(\(\frac{-x^{2}}{4}\)) = 2 ⇒ x³ = -8
x = -2
y = \(\frac{-x^{2}}{4}\) = – \(\frac{4}{4}\) = -1
Point of intersection is P(-2, -1)

= \(\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|\) = 3
Φ = tan^-1(3)

Question 24.
Find the angle between the curve 2y = e^{\(\frac{-x}{2}\)} and Y-axis.
Solution:
Equation of Y-axis is x = 0.
The point of intersection of the curve
2y = e^{\(\frac{-x}{2}\)} and x = 0 is P(0, \(\frac{1}{2}\))
Let ψ be the angle between the given curves
2y = e^{\(\frac{-x}{2}\)} at P with X – axis is given by
tan ψ = \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(0, \frac{1}{2}\right)}=\left.\frac{-1}{4} \mathrm{e}^{\frac{-x}{2}}\right|_{\left(0, \frac{1}{2}\right)}=\frac{-1}{4}\)
Further, if Φ is the angle between the Y – axis and 2y = e\(\frac{-x}{2}\), then we have
tan Φ = |tan (\(\frac{\pi}{2}\) – ψ)| = |cot ψ| = 4
∴ The angle between the curve and the Y-axis is tan^-1 4.

Question 25.
Show that the condition of the orthogonality of the curves ax² + by² = 1 and a₁x² + b₁y² = 1 is \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\).
Solution:
Let the curves ax² + by² = 1 and a₁x² + b₁y² = 1 intersect at p(x₁, y₁) so that
ax₁² + by₁² = 1 and a₁x₁² + b₁y₁² = 1, from which we get,
\(\frac{x_{1}^{2}}{b_{1}-b}\) = \(\frac{y_{1}^{2}}{a_{1}-a}\) = \(\frac{1}{a b_{1}-a_{1} b}\) …………. (1)
Differentiating ax² + by² = 1 with respect to x,
we get \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a x}{b y}\)
Hence, if mt is the slope of the tangent at P(x₁, y₁) to the curve
ax² + by² = 1, m₁ = \(\frac{-a x_{1}}{b y_{1}}\)
Similarly, the slope (m₂) of the tangent at P to
a₁x² + b₁y² = 1 is given by m₂ = \(\frac{-a_{1} x_{1}}{b_{1} y_{1}}\)
Since the curves cut orthogonally we have m₁m₂ = -1,
i.e., \(\frac{\mathrm{a} a_{1} x_{1}^{2}}{\mathrm{~b} \mathrm{~b}_{1} y_{1}^{2}}\) = -1 or \(\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{-b_{1}}{a a_{1}}\) ………………….. (2)
Now from (1) and (2), the condition for the orthogonality of the given curves is
\(\frac{b_{1}-b}{a-a_{1}}\) = \(\frac{b b_{1}}{a a_{1}}\)
or (b – a) a₁b₁ = (b₁ – a₁) ab
or \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\)

Question 26.
Show that the curves y² = 4(x + 1)and y² = 36 (9 – x) intersect orthogonally. [Mar 11, May 06, 05]
Solution:
Solving y² = 4(x + 1) and y² = 36 (9 – x) for the points of intersection, we get
4(x + 1) = 36 (9 – x) 10x = 80 or x = 8
i.e., y² = 4(x + 1) ⇒ y² = 4(9) = 36 ⇒ y = ± 6
The points of intersection of the two curves are P(8, 6), Q(8, -6)
y² = 4(x + 1) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{y}\)
y² = 36 (9 – x) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-18}{y}\)
Slope of the tangent to the curve
y² = 4(x + 1) at P is
m₁ = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Slope of the tangent to the curve
y² = 36 (9 – x) at P is
m₂ = \(\frac{-18}{6}\) = -3
m₁m₂ = \(\frac{1}{3}\) × -3 = -1
⇒ the curves intersect orthogonally at P.
We can prove, similarly,that the curves intersect orthogonally at Q also.

Question 27.
Find the average rate of change of s = f (t) = 2t² + 3 between t = 2 and t = 4.
Solution:
The average rate of change of s between t = 2 and t = 4 is \(\frac{f(4)-f(2)}{4-2}\) = \(\frac{35-11}{4-2}\) = 12.

Question 28.
Find the rate of change of area of a circle w.r.t. radius when r = 5 cm.
Solution:
Let A be the area of the circle with radius r.
Then A = πr². Now, the rate of change of area A w.r.t. r is given by \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2πr. When r = 5 cm.
\(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 10 π.
Thus, the area of the circle is changing at the rate of 10 π cm²/cm.

Question 29.
The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters ?
Solution:
Let x be the length of the edge of the cube, V be its volume and S be its surface area. Then, V = x³ and S = 6x². Given that rate of change of volume is 9 cm³/sec.
Therefore, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 9 cm³/sec.
Now differentiating V w.r.t. t, we get,

Question 30.
A particle is moving in a straight line so that after t seconds its distance is s (in cms) from a fixed point on the line is given by s = f(t) = 8t + t³. Find (i) the velocity at time t = 2 sec (ii) the initial velocity (iii) acceleration at t = 2 sec.
Solution:
The distance s and time t are connected by the relation.
s = f(t) = 8t + t³ …………. (1)
∴ velocity (ν) = 8 + 3t² ……………. (2)
and the acceleration is given by
a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 6t
i) The velocity at
t = 2 is 8 + 3 (4) = 20cm/sec.
ii) The initial velocity (t = 0) is 8 cm/sec.
iii) The acceleration at t = 2 is 6(2) = 12 cm/sec²

Question 31.
A container in the shape of an inverted cone has height 12 cm and radius 6 cm at the top. If it is filled with water at the rate of 12 cm³/sec., what is the rate of change in the height of water level when the tank is filled 8 cm ?
Solution:
Let OC be height to water level at t sec. The triangles OAB ad OCD are similar triangles.

Let OC = h and CD = r. Given that AB = 6 cm, OA = 12 cm.
\(\frac{r}{6}\) = \(\frac{h}{12}\)

Hence, the rate of change of water level is \(\frac{3}{4 \pi}\) cm/sec when the water level of the tank is 8 cm.

Question 32.
A particle is moving along a line according to s = f (t) = 4t³ – 3t² + 5t -1 where s is measured in meters and t is measured in seconds. Find the velocity and acceleration at time t. At what time ‘ the acceleration is zero.
Solution:
Since f(t) = 4t³ – 3t² + 5t – 1, the velocity at time t is
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t² – 6t + 5
and the acceleration at time t is a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 24t – 6.
The acceleration is 0 if 24t – 6 = 0
i.e., t = \(\frac{1}{4}\)
The acceleration of the particle is zero at
t = \(\frac{1}{4}\) sec.

Question 33.
The quantity (in mg) of a drug in the blood at time t (sec) is given by q = 3(0.4)^t. Find the instantaneous rate of change at t = 2 sec.
Solution:
Given that q = 3(0.4)^t
∴ \(\frac{\mathrm{dQ}}{\mathrm{dx}}\) = 3(0.4)^t log_e(0.4) is the instaneous rate of change in q. Hence the instaneous rate of q at time t = 2 sec. is given by
\(\left(\frac{\mathrm{dQ}}{\mathrm{dx}}\right)_{t=2}\) = 3(0.4)² log_e (0.4).

Question 34.
Let a kind of bacteria grow by t³ (t in sec). At what time the rate of growth of the bacteria is 300 bacteria per sec ?
Solution:
Let g be the amount of growth of bacteria at t sec. Then
g(t) = t³ ………… (1)
The growth rate at time t is given by
g'(t) = 3t²
300 = 3t² (given that growth rate is 300)
t = 10 sec.
∴ After t = 10 sec, the growth rate of bacteria should be 300 bacteria/sec.

Question 35.
The total cost C(x) in rupees associated with production of x units of an item is given by C(x) = 0.005 x³ – 0.02x² + 30x + 500. Find the marginal cost when 3 units are produced (marginal cost is the rate of change of total cost).
Solution:
Let M represent the marginal cost. Then
M = \(\frac{\mathrm{dC}}{\mathrm{dx}}\)
Hence,
M = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(0.005x³ – 0.02x² + 30x + 500) dx
= 0.005(3x²) – 0.02(2x) + 30
∴ The Marginal cost at x = 3 is
(M)_{x = 3} = 0.005 (27) – 0.02 (6) + 30 = 30.015.
Hence the required marginal cost is Rs. 30.02 to produce 3 units.

Question 36.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. Find the marginal revenue when x = 5 (marginal revenue is the rate of change of total revenue).
Solution:
Let m denote the marginal revenue. Then
m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) (since the total revenue is R(x))
Given that R(x) = 3x² + 36x + 5
∴ m = 6x + 36
The marginal revenue at x = 5 is
[m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\)]_{x = 5} = 30 + 36 = 66
Hence the required marginal revenue is Rs. 66.

Question 37.
Verify Rolle’s theorem for the function y = f(x) = x² + 4 in [-3, 3].
Solution:
Here f(x) = x² = 4. f is continuous on [-3, 3] as x² + 4 is a polynomial which is continuous on any closed interval. Further f(3) = f(-3) = 13 and f is differentiable on [-3, 3].
∴ By Rolle’s theorem ∃ c ∈ (-3, 3) such that f'(c) = 0
The point c = 0 ∈ (-3, 3). Thus Rolle’s theorem is verified.

Question 38.
Verify Rolle’s theorem for the function f(x) = x(x + 3)e^-x/2 in [-3, 0].
Solution:
Here f(-3) = 0 and f(0) = 0.
We have
f'(x) = \(\frac{\left(-x^{2}+x+6\right)}{2} e^{\frac{-x}{2}}\)
f'(x) = 0 ⇔ -x² + x + 6 = 0 ⇔ x = -2 or 3. Of these two values -2 is in the open interval (-3, 0) which satisfies the conclusion of Rolle’s theorem.

Question 39.
Let f(x) = (x -1) (x – 2) (x – 3). Prove that there is more than one ‘c’ in (1, 3) such that f'(c) = 0. [Mar 13]
Solution:
Observe that f is continuous on (1, 3) differentiable in (1, 3) and f(1) = f(3) = 0.
f'(x) = (x – 1) (x – 2) + (x – 1) (x – 3) + (x – 2) (x – 3)
= 3x² – 12x + 11
\(\frac{12 \pm \sqrt{144-132}}{6}\) = 2 ± \(\frac{1}{\sqrt{3}}\)
Both these roots lie in the open interval (1, 3) and are such that the derivative vanishes at these points.

Question 40.
On the curve y = x², find a point at which the tangent is parallel to the chord joining (0, 0) and (1, 1).
Solution:
The slope of the chord is \(\frac{1-0}{1-0}\) = 1.
The derivative is \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x.
We want x such that 2x = 1
i.e., x = \(\frac{1}{2}\)
we not that \(\frac{1}{2}\) is in the open interval (0, 1), as required in the Lagrange’s mean value theorem.
The corresponding point on the curve is (\(\frac{1}{2}\), \(\frac{1}{4}\)).

Question 41.
Show that f(x) = 8x + 2 is a strictly increasing function on R without using the graph of y = f(x).
Solution:
Let x₁, x₂ ∈ R with x₁ < x₂. Then 8x₁ < 8x₂. Adding 2 to both sides of this inequality, we have 8x₁ + 2 < 8x₂ + 2. i.e., f(x₁) < f(x₂).
Thus
x₁ < x₂ ⇒ f(x₁) < f(x₂) ∀ x₁, x₂ ∈ R.
Therefore, the given function f is strictly increasing on R.

Question 42.
Show that f (x) = e^x is strictly increasing on R (without graph).
Solution:
Let x₁, x₂ ∈ R such that x₁ < x₂. we know that if a > b then e^a > e^b
∴ x₁ < x₂ ⇒ e^x₁ < e^x₂
i.e., f(x₁) < f(x₂).
Hence the given function f is a strictly increasing function.

Question 43.
Show that f(x) = – x + 2 is strictly decreasing on R.
Solution:
Let x₁, x₂ ∈ R x₁ < x₂.
Then x₁ < x₂
⇒ -x₁ > -x₂
⇒ -x₁ + 2 > -x₂ + 2
⇒ f(x₁) > f(x₂).
Therefore the given function f is strictly decreasing on R.

Question 44.
Find the intervals on which
f(x) = x² – 3x + 8 is increasing or decreasing ?
Solution:
Given fucntion is f(x) = x² – 3x + 8.
Differentiating it w.r.t. x, we get f'(x) = 2x – 3
f(x) = 0 for x = 3/2.

since f'(x) < 0 in (-∞, 3/2) the function f(x) is strictly decreasing on (-∞, \(\frac{3}{2}\)) Further since f'(x) > 0 in (\(\frac{3}{2}\), -∞), the function f(x) is a strictly increasing function (\(\frac{3}{2}\), -∞).

Question 45.
Show that f(x) = |x| is strictly decreasing on (-∞, 0) and strictly increasing on (0, ∞).
Solution:
The given function is f(x) = |x| i.e.,

Thus f'(c) = 1 if c > 0, f'(c) = -1 if c < 0. Since f'(c) > 0 on (0, ∞), the function f(x) is strictly increasing on (0, ∞). Since f'(c) < 0 on (-∞, 0), the function f(x) is strictly decreasing on (-∞, 0).

Question 46.
Find the intervals on which the function f(x) = x³ + 5x² – 8x + 1 is a strictly increasing function.
Solution:
Given that f(x) = x³ + 5x² – 8x + 1.
∴ f'(x) = 3x² + 10x – 8 = (3x – 2) (x + 4)
= 3(x – \(\frac{2}{3}\)) (x – (-4)).
f'(x) is negative in (-4, \(\frac{2}{3}\)) and positive in (-∞, -4) ∪ (\(\frac{2}{3}\), ∞) .
∴ The function is strictly deceasing in (-4, \(\frac{2}{3}\)) and is strictly decreasing in (-∞, -4)and (\(\frac{2}{3}\), ∞)

Question 47.
Find the Intervals on which f(x) = x^x (x > 0) is increasing and decreasing.
Solution:
Taking logarithms on both sides of f(x) = x^x
we get
log (f(x)) = x log x. Differenetiating it w.r.t. x
we have \(\frac{1}{f(x)}\) f'(x)= 1 + log x
∴ f'(x) = x^x( 1 + log x)
f'(x) = 0 ⇒ x^x(1 + log x) = 0 ……………. (1)
⇒ 1 + log x = 0
⇒ x = 1/e

Suppose x < 1/e log x < log (1/e) (since the base e > 1). i.e., log x < -1
1 + log x < 0 ⇒ x^x (1 + log x) < 0. i.e., f'(x) < 0 Now suppose, x > 1/e. Then log x > log (1/e)
i.e., log x > – 1.
⇒ 1 + log x > 0
⇒ x^x (1 + log x) > 0
⇒ f'(x) > 0
Hence, f is strictly decreasing on (o, 1/e) and it is strictly increasin on (1/e, ∞).

Question 48.
Determine the intervals in which f(x) = \(\frac{2}{(x-1)}\) + 18x ∀ x ∈ R \ {0} is strictly increasing and decreasing.
Solution:
Given that f(x) = \(\frac{2}{(x-1)}\) + 18x. Differenetiating
it w.r.t. x, we get
f'(x) = \(\frac{-1}{(x-1)^{2}}\) . 2 + 18 and f'(x) = 0
⇒ \(\frac{2}{(x-1)^{2}}\) = 18 ⇒ (x – 1)² = 1/9.
∴ f'(x) = 0 if x – 1 = 1/3 or x- 1 = -(1/3).
i.e., x = 4/3 or x – 2/3.
The derivative of f(x) can be expressed as
f'(x) = \(\frac{18}{(x-1)^{2}}\) . (x – /3) (x – 4/3)

∴ The given function f(x) is strictly increasing on (-∞, \(\frac{2}{3}\)) and (\(\frac{4}{3}\), ∞) and it is strictly decreasing on (\(\frac{2}{3}\), \(\frac{4}{3}\)).

Question 49.
Let f(x) = sin x – cos x be defined on [0, 2π]. Determine the intervals in which f(x) is strictly decreasing and strictly increasing.
Solution:
Given that f(x) = sin x – cos x.
∴ f'(x) = cos x + sin x
∴ f'(x) = \(\sqrt{2}\) . sin(x + π/4)
Let 0 < x < 3π/4. Then π/4 < x + π/4 < π. ∴ sin (x + π/4) > 0 i.e., f'(x) > 0.
Similarly it can be shown that f'(x) < 0 in (3π/4 . 7π/4) and f'(x) > 0 in (7π/4, 2π).

Thus the function f(x) strictly increasing in (0, \(\frac{3\pi}{24}\) and (\(\frac{7\pi}{4}\), 2π) it is strictly decreasing in (\(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\)).

Question 50.
If 0 ≤ x ≤ \(\frac{\pi}{2}\) then show that x ≥ sinx.
Solution:
Let f(x) = x – sin x. Then f'(x) = 1 – cos x ≥ 0 ∀ x
∴ f is an increasing function for all x.
Now, f(0) = 0. Hence f(x) ≥ f(0) for all x ∈ (0, \(\frac{\pi}{2}\)). Therefore, x ≥ in x.

Question 51.
Let f : R → R be defined by f(x) = 4x² – 4x + 11. Find the global minimum value and a point of global minimum.
Solution:
We have to look for a value c e R(domain) such that
f(x) ≥ f(c) ∀ x ∈ R
so that f(c) is the global minimum value of f. Consider
f(x) = 4x² – 4x + 11 = (2x – 1)² + 10 ≥ ∀ x ∈ R ……………(1)
Now, f(1/2) = 10
Also f(x) ≥ f(1/2) ∀ x ∈ R
Hence, f(1/2) = 10 is the global minimum value of f(x), and a point of global minimum is x = 1/2.

Question 52.
Let f : [-2, 2] → R be defined by f(x) = |x|. Find the global maximum of f(x) and a point of global minimum.
Solution:
We know that |x| = \(\left\{\begin{array}{ccc}
x & \text { if } & x \geq 0 \\
-x & \text { if } & x<0
\end{array}\right.\)
Therefore, from the graph of the function f on [-2, 2] clearly f(x) ≤ f(2) and f(x) ≤ f(-2) ∀ x ∈ [-2, 2].

∴ f(2) = f(-2) = 2 is the global maximum of f(x), 2 and -2 are the points of global maximum.

Question 53.
Find the global maximum and global minimum of the function f : R → R defined by f (x) = x².
Solution:
We have f(x) ≥ f(0) ∀ x ∈ R.
Hence th global minimum value of f(x) is 0 and a point of global minimum is x = 0.
Suppose f has global maximum at x₀ ∈ R (x₀ > 0). Then as per out assumption we have.
f(x₀) ≥ f(x) ∀ x ∈ R

Choose x₁ = x₀ + 1. Then x₁ ∈ R and x₀ < x₁
∴ x₀² < x₁²
Hence f(x₀) < (fx₁)
Thus we got f(x₁) such that f(x₀) > f(x₀) which is a contradition to Therefore, f(x) has no global maximum on R.

Question 54.
Find the stationary points of f(x) = 3x⁴ – 4x³ + 1, ∀ x ∈ R and state whether the function has local maxima or local minima at those points.
Solution:
Given that f(x) = 3x⁴ – 4x³ +1 and the domain of f is R. Differentiating the function w.r.t. x we have
f'(x) = 12x²(x – 1) ……………………… (1)
The stationary points are the roots of f (x) = 0 i.e., 12x²(x – 1) = 0. Hence x = 0 and x = 1 are the stationary points. Now, we test whether the stationary point x = 1 is a local extreme point or not. For.
f'(0.9) = 12(0.9)² (0.9 – 1) ⇒ f'(0.9) is negative
f'(1.1) = 12(1.1)² (1.1 – 1) ⇒ f'(1.1) is positive
and f(x) is defined in the neighbourhood i.e., (0.8, 1.2) of x = 1 with 8 = 0.2.
By theorem cis a point of local maximum if f(x) changes sign from positive to negative at x = c.
c is a point of local minimum if f'(x) changes sign from negative to positive at x = c.
The given function has local (relative) minimum at x = 1. Hence x = 1 is a local extreme point.
We will now test whether x = 0 is a local extreme point or not.
The function f(x) is defined in the neighbourhood of (-0.2, 0.2).
f'(-0.1) = 12(-0.1)² (-0.1 – 1) ⇒ f'(-0.1) is negative
f'(-0.1) = 12(0.1)² (0.1 -1) ⇒ f'(0.1) is negative
Thus, f(x) has no change in sign at x = 0. Therefore, the function f has no local maximum and no local minimum. Hence, x = 0 is not a local extreme point.

Question 55.
Find the points (if any) of local maxima and local minima of the function f (x) = x³ – 6x² + 12x – 8 ∀ x ∈ R.
Solution:
Given function is f(x) = x³ – 6x² + 12x – 8 and the domain of f is R.
Differentiating the given function w.r.t. x, we get
f'(x) = 3x² – 12x + 12 i.e., f'(x) = 3(x – 2)².
The stationary point of f(x) is x = 2, since 2 is a root of f'(x) = 0.
Choose δ = 0.2 The 0.2- neighbourhood of 2 is (1.8, 2.2). Now
f'(1.9) = 3(1.9 – 2)² ⇒ f'(1.9) is positive
f'(2.1) = 3(2.1- 2)² ⇒ f'(2.1) is positive
Thus f(x) does not change the sign at x = 2. By Theorem c is neither a point of local maximum nor a point of local minimum if f'(x) does not change sign at x = c.
x = 2 is neithere a local maximum nor a local minimum.

Question 56.
Find the points of local minimum and local maximum of the function f(x) = sin 2x ∀ x ∈ [0, 2π]
Solution:
The given function is f(x) = sin 2x and domain is [0, 2π].
f'(x) = 2cos 2x …………… (1)
The critical points are the roots of 2 cos 2x = 0 and lying in the domain [0, 2π].
They are \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) .
Now we apply the first derivative test at x = \(\frac{\pi}{4}\)
Clearly (\(\frac{\pi}{4}\) – 0.1 . \(\frac{\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{\pi}{4}\) and the given f is defined on it.
Now
f'(\(\frac{\pi}{4}\) – 0.05) = 2 cos(\(\frac{\pi}{2}\) – 0.1) > 0
f'(\(\frac{\pi}{4}\) + 0.05) = 2 cos(\(\frac{\pi}{2}\) + 0.1) < 0
Thus f'(x) changes sign from positive to negative at x = \(\frac{\pi}{4}\). Therefore f has a local maximum.
Now we apply the first derivative test at x = \(\frac{3\pi}{4}\).
Clearly (\(\frac{3\pi}{4}\) – 0.1 . \(\frac{3\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{3\pi}{4}\) and the given f is defined on it.
Now
f'(\(\frac{3\pi}{4}\) – 0.05) = 2 cos(\(\frac{3\pi}{4}\) – 0.1) < 0 f'(\(\frac{3\pi}{4}\) + 0.05) = 2 cos(\(\frac{3\pi}{4}\) + 0.1) > 0
Thus f'(x) changes sign from positive to negative at x = \(\frac{3\pi}{4}\). Therefore f has a local maximum at x = \(\frac{3\pi}{4}\).

Question 57.
Find the points of local extrema of the function f(x) = x³ – 9x² – 48x + 6 ∀ x ∈ R Also find its local extrema.
Solution:
Given function is
f(x) = x³ – 9x² – 48x + 6 …………… (1)
and the domain of the function is R.
Differentiating (1) w.r.t. x we get
f'(x) = 3x² – 18x – 48 = 3(x – 8) (x + 2) ………….. (2)
Thus the stationary points are – 2 and 8.
Differentiating (2) w.r.t.x we get,
f'(x) = 6(x – 3) ………….. (3)
Let x₁ = -2 and x₂ = 8. Now we have to find f’ at each of these points to know the sign of second derivative.
At x₁ = -2, f'(-2) = – 30. The sign of it is negative.
∴ x₁ = – 2 is a point of local maximum of f and its local maximum value is f(-2) – 58.
Now, at x₂ = 8 f'(8) = 30. Thus the sign of f”(x₂) is positive. Therefore, x = 8 is a point of local minimum of f and its local minimum value is
f(8) = – 442.

Question 58.
Find the points of local extrema of f(x) = x⁶ ∀ x ∈ R. Also find its local extrema.
Solution:
f(x) = x⁶ ………………….. (1)
Differentiating (1) w.r.t. x we get,
f'(x) = 6x⁵ ……………. (2)
and again differentiating (2) w.r.t. x we get
f'(x) = 30x⁴ …………………… (3)
The stationary point of f(x) is x = 0 only (since f'(x) = 0 only at x = 0).
Now f'(0) = 0. At x = 0, we can not conclude anything about the local extrema by the second derivative test. Therefore, we apply the first derivative test. As the domain of f is R, the function f is defined on (-0.2, 0.2) which is a neighbourhood of x = 0. Now
f'(-0.1) = 6(-0.1)5 < 0, f'(0, 1) = 6(0.1) 5 > 0.
Thus f'(x) changes sign form negative of positive at x = 0.
∴ x = 0 is a point of local minimum and its local minimum value is f(0) = 0.

Question 59.
Find the points of local extrema and local extrema for the function f(x) = cos 4x defined on (0, \(\frac{\pi}{2}\))
Solution:
Here f(x) = cos 4x ……………. (1)
and its domain is (0, \(\frac{\pi}{2}\))
∴ f’(x) = -4 sin 4x ………….. (2)
and f”(x) = -16 cos 4x ……………… (3)
The stationary points are the roots of
f'(x) = 0 and lying in the domain (o, \(\frac{\pi}{2}\)).
f'(x) = 0 ⇒ 4 sin 4x = 0
⇒ 4x = 0, π, 2π, 3π, 4π ………………….
⇒ x = 0, π/4, π/2, 3π/4, π …………………..
The point lying in the domain is x = \(\frac{\pi}{4}\) only.
Thus x = \(\frac{\pi}{4}\) is the stationary point of the given function. Now
f”(\(\frac{\pi}{4}\)) = -16 cos(π) = 16 > 0.
The function f has local minimum at x = \(\frac{\pi}{4}\) and its local minimum value is
f(\(\frac{\pi}{4}\)) = -1.

Question 60.
Find two positive number whose sum is 15 so that the sum of their squares is minimum.
Solution:
Suppose one numbers is x and the other number 15 – x. Let S be the sum of squares of these numbers. Then S = x² + (15 – x)² ………………… (1)
Note that the quantity S, to be minimized, is a function of x.
Differentiating (1) w.r.t. x, we get
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2x + 2(15 – x) (-1)
= 4x – 30 ………………. (2)
and again differentiating (2) w.r.t.x, we get
\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 ……………….. (3)
The stationary point can be obtained by solving \(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0 i.e., 4x – 30 = 0.
∴ x = 15/2 is the stationary point of (1).
Since \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 > 0, S is minimum at x = \(\frac{15}{2}\)
∴ The two numbers are \(\frac{15}{2}\), 15 – \(\frac{15}{2}\) i.e., \(\frac{15}{2}\) and \(\frac{15}{2}\).

Question 61.
Find the maximum area of the rectangle that can be formed with fixed perimeter 20.
Solution:
Let x and y denote the length and the breadth of a rectangle respectively. Given that the perimeter of the rectangle is 20.
i.e., 2(x + y) = 20
i.e., x + y = 10 ……………. (1)
Let A denote the area of rectangle.
Then A = xy ………….. (2)
Which is to be minimized. Equation (1) can be expressed as
y = 10 – x …………… (3)
From (3) and (2), we have
A = x (10 – x)
A = 10x – x² ……………… (4)
Differentiating (4) w.r.t. x we get
\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 10 – 2x ……………….. (5)
The stationary point is a root of 10 – 2x = 0
∴ x = 5 is the stationary point.
Differentiating (5) w.r.t. x, we get
\(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\) = -2
which is negative. Therefore by second derivative test the area A is maximized at x = 5 and hence y = 10 – 5 = 5, and the maximum area is A = 5(5) = 25.

Question 62.
Find the point on the graph y² = x which is the nearest to the point (4, 0).
Solution:

Let P(x, y) be any point on y² = x and A(4, 0). We have to find P such that PA is minimum
Suppose PA = D. The quantity to be minimized is D.
D = (\(\sqrt{(x-4)^{2}+(y-0)^{2}}\)) ………………. (1)
P(x, y) lies on the curve, therefore
y² = x …………………… (2)
From (1) and (2), we have
D = \(\sqrt{\left((x-4)^{2}+x\right)}\)
D = \(\sqrt{\left(x^{2}-7 x+16\right)}\) ……………………. (3)
Differentiating (3) w.r.t. x, we get
\(\frac{\mathrm{dD}}{\mathrm{dx}}\) = \(=\frac{2 x-7}{2} \cdot \frac{1}{\sqrt{x^{2}-7 x+16}}\)
Now \(\frac{\mathrm{dD}}{\mathrm{dx}}\) = 0
gives x = \(\frac{7}{2}\). Thus \(\frac{7}{2}\) is a stationary point of the function D. We apply the first derivative test to verify whether D is minimum at x = \(\frac{1}{2}\)

Question 63.
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Solution:
Let O be the centre of the circular base of the cone and its height be h. Let r be the radius of the circular base of the cone.
Then AO = h, OC = r.
Let a cylinder with radius x(OE) be inscribed in the given cone. Let its height be u.

i.e., RO = QE = PD = u
Now the triangles AOC and QEC are similar.
Therefore,
\(\frac{\mathrm{QE}}{\mathrm{OA}}\) = \(\frac{\mathrm{EC}}{\mathrm{OC}}\)
i.e., \(\frac{u}{h}\) = \(\frac{r-x}{r}\)
∴ u = \(\frac{h(r-x)}{r}\) ………………… (1)
Let S denote the curved surface area of the chosen cylinder.Then
S = 2 π xu.
As the cone is fixed one, the values of r and h are constants. Thus S is function of x only. Now,
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2 πh (r – 2x)/r and \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = -4πh/r.
The stationary point of S is a root of
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0
i.e., π(r – 2x)/r = 0
i.e., x = r/2
\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) < 0 for all x, Therefore (\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\))_x=r/2 < 0.
Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is r/2.

Question 64.
The profit function P(x) of a company, selling x items per day is given by P(x) = (150 – x)x – 1600. Find the number of items that the company should sell to get maximum profit. Also find the maximum profit.
Solution:
Given that the profit function is
P(x) = ( 150 – x)x – 1600 ………….. (1)
For maxima or minima \(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{dx}}\) = 0
∴ (150 – x) (1) + x (-1) = 0
i.e., x = 75
Now \(\frac{d^{2} P(x)}{d x^{2}}\) = -2 and \(\left[\frac{d^{2} P(x)}{d x^{2}}\right]_{x=75}\) < 0.
∴ The profit P(x) is maximum for x = 75.
∴ The company should sell 75 items a day to make maximum profit.
The maximum profit will be P(75) = 4025.

Question 65.
A manufacturer can sell x items at a price of rupees (5 – x/100) each. The cost price of x items is Rs. (x/5 + 500). Find the number of items that the manufacturer should sell to earn maximum profits.
Solution:
Let S(x) be the selling price of x items and C(x) be the cost price of x items. Then, we have
S(x) = {cost of each item}, x .
∴ S(x) = (5 – x/100) x = 5x – x²/100 and C(x) = x/5 + 500
Let P(x) denote the profit function. Then
P(x) = S(x) – C(x)
P(x) = (5x – x²/100) – (x/5 + 500)
– (24x/5) – (x²/100) – 500 …………….. (1)
For maxima or minima
\(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{d} x}\) = 0
i.e., 24/5 – x/50 = 0
The stationary point of P(x) is x = 240 and
\(\left[\frac{d^{2} P(x)}{d x^{2}}\right]\) = –\(\frac{1}{50}\) for all x.
Hence the manufacturer can earn maximum profit if he sells 240 items.

Question 66.
Find the absolute extrement of f(x) = x² defined on [-2, 2].
Solution:
The given function f(x) = x² is continuous on [-2, 2]. It can be shown that it has only local minimum and the point of local minimum is 0. The absolute(global) maximum of f is the largest value of f(-2), f(0) and f(2) i.e., 4, 0, 4.
Hence, the absolute maximum value is 4. Similarly the absolute minimum is the least value of 4, 0, 4. Hence 0 is the absolute minimum value.

Question 67.
Find the absolute maximum of x⁴⁰ – x²⁰ on the interval [0, 1]. Find also its absolute maximum value.
Solution:
Let f(x) = x⁴⁰ – x²⁰ ∀ x ∈ [0,1] ………… (1)
The function f is continuous on [0,1 j and the interval [0,1] is closed.
From (1) we have
f'(x) = 40 x³⁹ – 20 x¹⁹ = 20x¹⁹ (2x²⁰ – 1).
Thus f'(x) = 0 at x = 0 or
x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)
Therefore, the critical points of f are and \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\) and 0 is one of the end points of the domain. Therefore no local maximum exists at x = 0. Now
f'(x) = 40(39) x³⁸ – 20(19) x¹⁸
= 20x¹⁸ (78 x²⁰ – 19)
[f”(x)]_{x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)} = 20 (1/2)^(18/20)[39 – 19] > 0.
Therefore f has local minimum at
x = (1/2)(1/20)
and its value is \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) = –\(\frac{1}{4}\)
Therefore the absolute maximum value of the function f is the largest value of f(0), f(1) and \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) i.e., the largest value of {0, 0, –\(\frac{1}{4}\)}
Hence, the absolute maximum of f is 0 and the points of absolute maximum are 0 and 1. Further the absolute minimum is the least of 0, 0, -latex]\frac{1}{4}[/latex].
Hence the absolute minimum is -latex]\frac{1}{4}[/latex] and the point of absolute minimum is x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Differentiation Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Differentiation Important Questions

Question 1.
If f(x) = x² (x ∈ R), prove that f is differentiable on R and find its derivative.
Solution:
Given that f(x) = x²
for x, h ∈ R, f(x + h) – f(x)(x + h)² – x²
= x² + h² + 2hx – x²
= 2hx + h² = h(2x + h)

∴ f is differentiable on R and f'(x) = 2x for each x ∈ R

Question 2.
Suppose f(x) = \(\sqrt{x}\) (x > 0). Prove that f is differentiable on (0, ∞) and find f(x).
Solution:
Let x ∈ (0, ∞) h ≠ 0 and |h| < 0

Question 3.
If f(x) = \(\frac{1}{x^{2}+1}\) (x ∈ R), prove that f is differentiable on R and find f'(x).
Solution:
Let x ∈ R and h ≠ 0

= –\(\frac{2 x}{\left(x^{2}+1\right)^{2}}\)
∴ f is differentiable and f'(x) = –\(\frac{2 x}{\left(x^{2}+1\right)^{2}}\) for each x ∈ R .

Question 4.
If f(x) = sin x (x ∈ R), then show that f is differentiable on R and f'(x) = cosx.
Solution:
Let x ∈ R and h ≠ 0

∴ f is differentiable on R and f'(x) = cos x for each x ∈ R.

Question 5.
Show that f(x) = |x| (x ∈ R) is not differentiable at zero and is differentiable at any x ≠ 0.
Solution:
Given f(x) = |x|
∴ f(x) = x if x ≥ 0
if h ≠ 0
\(\frac{f(0+h)-f(0)}{h}\) = \(\frac{f(h)}{h}\) = \(\left\{\begin{array}{r}
1 \text { if } h>0 \\
-1 \text { if } h<0
\end{array}\right.\)
f'(0⁺) = 1, f'(0^–) = -1
∴ is not differentiable at zero it can be easily proved that f is differentiable at any x ≠ 0 and that f'(x) = \(\left\{\begin{array}{l}
1 \quad \text { if } x>0 \\
-1 \text { if } x<0
\end{array}\right.\)

Question 6.
Check whether the following function is differentiable at zero f(x) = \(\left\{\begin{array}{l}
3+x \text { if } x \geq 0 \\
3-x \text { if } x<0
\end{array}\right.\)
Solution:

f has the left hand derivative at zero and f'(0^–) = -1
∴ f'(0⁺) ≠ f'(0^–)
f(x) is not differentiable at zero.

Question 7.
Show that the derivative of a constant function on an interval is zero.
Solution:
let f be a constant function on an interval I.
f(x) = C ∀ x ∈ I for some constant.
Let a ∈ I, for h ≠ 0 \(\frac{f(a+h)-f(a)}{h}\) = \(\frac{c-c}{h}\) = 0
for sufficiently small (h)

∴ f is differentiable 0 and f'(0).

Question 8.
Suppose for all x, y ∈ R f(x + y) = f(x). f(y) and f'(0) exists. Then show that f(x) exists and equals to f(x) f'(0)for all x ∈ R.
Solution:
Let x ∈ R, for h ≠ 0, we have
\(\frac{f(x+h)-f(x)}{h}\) = \(\frac{f(x) f(h)-f(x)}{h}\)
= f(x) \(\frac{[f(h)-1]}{h}\) ………………….. (1)
f(0) = f(0 + 0) = f(0) f(0) ⇒ f(0) (1 —f(0)) = 0
∴ f(0) = 0 or f(0) = 1
Case (1) : Suppose f(0) = 0
f(x) = f(x + 0) = f(x) f(0) = 0 ∀ x ∈ R
∴ f(x) is a constant function = f'(x) = 0 for all x ∈ R
∴ f'(x) = 0 = f(x) . f'(0)

Case (ii): Suppose f(0) = 1

∴ f is differentiable and f'(x) = f'(x) f'(0).

Question 9.
If f(x) = (ax + b)ⁿ, (x > –\(\frac{b}{a}\)), then find f'(x).
Sol:
Let u = ax + b so that y = uⁿ
f'(x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uⁿ) \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= n.u^n-1a.
= an(ax+b)^{n – 1}

Question 10.
Find the derivative of f(x) = e^x (x² + 1)
Solution:
Let u = e^x, V = x² + 1
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = e^x, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 2x
f'(x) = u(x) v'(x) + u'(x) . v(x)
= e^x2x + (x² + 1) e^x
= e^x(2x + x² + 1)
= e^x(x + 1)²

Question 11.
If y = \(\frac{a-x}{a+x}\) (x ≠ -a), find \(\frac{\mathrm{d} y}{\mathrm{dx}}\).
Solution:
Let u = a – x and v = a + x so that y = \(\frac{\mathrm{u}}{\mathrm{v}}\)

Question 12.
If f(x) = e^2x . log x (x > 0), then find f'(x).
Solution:
Let u = e^2x, v = log x so that
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2 . e^2x , \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 1
f(x) = u.v
f'(x) = u . \(\frac{\mathrm{dv}}{\mathrm{dx}}\) + y . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= e^2x . \(\frac{1}{x}\) + log x (2e^2x)
= e^2x (\(\frac{1}{x}\) + 2 logx)

Question 13.
If f(x) = \(\sqrt{\frac{1+x^{2}}{1-x^{2}}}\) (|x| < 1), then find f'(x)
Solution:

Question 14.
If f(x) = x², 2^x log x (x > 0), find f'(x).
Solution:
u = x², v = 2^x, w = logx
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2x, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 2^x . log2, \(\frac{\mathrm{dw}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
f'(x) = uv . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) + vw . \(\frac{\mathrm{du}}{\mathrm{dx}}\) + uw . \(\frac{\mathrm{dv}}{\mathrm{dx}}\)
= x²2^x . \(\frac{1}{x}\) + 2^x . logx(2x) + x² . logx . 2^x log2
= x . 2^x (logx² + xlogx (log 2) + 1)

Question 15.
If y = \(\left|\begin{array}{l}
f(x) g(x) \\
\phi(x) \psi(x)
\end{array}\right|\) then show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\left|\begin{array}{l}
f^{\prime}(x) g^{\prime}(x) \\
\phi(x) \psi(x)
\end{array}\right|\) + \(\left|\begin{array}{l}
f(x) g \backslash(x) \\
\phi^{\prime}(x) \Psi(x)
\end{array}\right|\)
Solution:
Given y = \(\left|\begin{array}{l}
f(x) g(x) \\
\phi(x) \psi(x)
\end{array}\right|\)
= f(x) ψ(x) – Φ(x) g(x)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = f(x) ψ'(x) + ψ(x) f'(x) – [Φ(x). g'(x) + g(x). Φ'(x)]
= [f(x) ψ'(x) – g(x) Φ’x)] + [f'(x) ψ(x) – Φ(x).g'(x)]
= \(\left|\begin{array}{l}
f(x) g(x) \\
\phi^{\prime}(x) \psi^{\prime}(x)
\end{array}\right|\) + \(\left|\begin{array}{cc}
f^{\prime}(x) & g^{\prime}(x) \\
\phi(x) & \psi(x)
\end{array}\right|\)

Question 16.
If f(x) = 7^x2+3x (x > 0), then find f'(x).
Solution:
Let u = x³ + 3x ⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 3x² + 3 = 3(x² + 1)
f(X) = 7^u
f'(x) = \(\frac{d f}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\) (7^u . l0g 7) [3(x² + 1)]
= 3(x² + 1) 7^x2+3x log 7

Question 17.
If f(x) = x e^x sin x, then find f(x).
Solution:
Let u = x ,v = e^x, w = sin x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = e^x . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) = cos x
f(x) = u.v.w
f'(x) = uv . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) + uw \(\frac{\mathrm{dv}}{\mathrm{dx}}\) + vw \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= xe^x cos x + x . sinx e^x + e^x sin ^x.

Question 18.
If f(x) = sin (log x), (x > 0), find f'(x).
Solution:
Let u = logx, y = f(x) so that y = sin u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos u, \(\frac{\mathrm{du}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
f'(x) = \(\frac{1}{x}\) . cos u = \(\frac{1}{x}\) cos (log x)

Question 19.
If f(x) =(x³ + 6x² + 12x – 13)¹⁰⁰; find f'(x).
Solution:
u = x³ + 6x² + 12x – 13
⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 3x² + 12x + 12
= 3(x² + 4x + 4)
= 3(x + 2)²
f(x) = u¹⁰⁰
f'(x) = 100 . u⁹⁹ . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= 100 (x³ + 6x² + 12x – 13)⁹⁹ . 3(x + 2)²
= 300 (x + 2)² (x³ + 6x² + 12x – 13)⁹⁹

Question 20.
Find the derivative of f(x) = \(\frac{x \cos x}{\sqrt{1+x^{2}}}\)
Solution:
Let u = x cos x, and v = \(\sqrt{1+x^{2}}\) so that

Question 21.
li f(x) = log (secx + tan x), find f'(x). [Mar 14, May 11]
Solution:
Let u = sec x + tan x and y = log u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\), \(\frac{\mathrm{du}}{\mathrm{dx}}\) = sec x. tan x + sec² x
= sec x (sec x + tan x)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= \(\frac{1}{\sec x+\tan x}\) . sec x(sec x + tan x) = sec x

Question 22.
If y = sin^-1\(\sqrt{x}\), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Solution:
u = \(\sqrt{x}\), y = sin^-1 x.

Question 23.
If y = sec (\(\sqrt{\tan x}\)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = \(\sqrt{\tan x}\), v = tanx
Then y = sec u, u = \(\sqrt{\mathrm{v}}\), v = tan x

Question 24.
If y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let u = x sin^-1x, v = \(\sqrt{1-x^{2}}\)

Question 25.
If y = log (cosh 2x), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let u = cosh 2x, so that y = log u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\); \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2 sin h2x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= 2 sin h 2x . \(\cosh 2 x\) = 2 tan h 2x

Question 26.
If y = log (sin (log x)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let v = log x, u = sin v so that y = log u.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\); \(\frac{d u}{d v}\) = cos u; \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{d u}{d v}\) . \(\frac{\mathrm{dv}}{\mathrm{dx}}\)
= \(\frac{1}{\sin (\log x)}\) . cos (logx) \(\frac{1}{x}\) = \(\frac{\cot (\log x)}{x}\)

Question 27.
If y = (cot^-1x³)², find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = cot^-1x³, u = x³, y = u²

Question 28.
If y = cosec^-1(e^2x+1), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = e^2x+1, y = cosec^-1u

Question 29.
If y = tan^-1 (cos \(\sqrt{x}\)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
v = \(\sqrt{x}\) and u = cos v, y = tan^-1u
\(\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{1}{2 \sqrt{x}}\), \(\frac{d u}{d v}\) = – sin u; \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{1+u^{2}}\)
= – sin \(\sqrt{x}\) = \(\frac{1}{1+\cos ^{2}(\sqrt{x})}\)

Question 30.
If y = Tan^-1 \(\sqrt{42}\) for 0 < |x| < 1, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\). [May, Mar 12]
Solution:
Put x² = cos 2θ

Question 31.
If y = x²e^xsin x, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
log y = log x². e^x. sin x
= log x² + log e^x + log sin x
= 2 log x + log e^x + log sin x
Differentiating w.r.to by sin x
\(\frac{1}{y}\) . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{x}\) + 1 + \(\frac{1}{sin x}\) . cos x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y(\(\frac{2}{x}\) + 1 + cot x)

Question 32.
If y = x^tanx + (sin x)^{cos x}, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [Mar. 14, 11]
Solution:
Let u = x^tanx and v = (sin x)^{cos x}
log u logx ^tanx = (tan x) log x

Question 33.
If x = a(cos t + log tan (\(\frac{t}{2}\))), y = a sin t, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:

Question 34.
If x^y = e^x-y, than show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\log x}{(1+\log x)^{2}}\)    [May 07]
Solution:
x^y = e^x-y
log x^y = log e^x-y
y log x = (x – y) (log e = 1)
y(1 + log x) = x

Question 35.
If siny = x sin (a + y), then show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\sin ^{2}(a+y)}{\sin a}\) (a is not a multiple of π)
Solution:

Question 36.
If y = x⁴ + tan x, then find y”.
Solution:
y = x⁴ + tan x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4x³ + sec² x
\(\frac{d^{2} y}{d x^{2}}\) = 12x² + 2 sec x (sec x tan x)
= 12x² + 2 sec²x . tan x

Question 37.
If f(x) = sinx, sin 2x sin 3x, find f”(x).
Solution:
f(x) = \(\frac{1}{2}\) sin 2x(2 sin 3x sin x)
= \(\frac{1}{2}\) (sin 2x) (cos 2x – cos4x)
= \(\frac{1}{4}\) (2 sin 2x cos 2x – 2 sin 2x cos 4x)
= \(\frac{1}{4}\) (sin2x + sin4x – sin6x)
Therefore,
f'(x)= \(\frac{1}{4}\)[2 cos 2x+ 4cos 4x – 6cos 6x]
Hence,
f”(x) = \(\frac{1}{4}\) (-4 sin 2x – 16 sin 4x + 36 sin 6x)
= 9 sin 6x – 4 sin 4x – sin 2x.

Question 38.
Show that y = x + tan x satisfies cos2x \(\frac{d^{2} y}{d x^{2}}\) + 2x = 2y.
Solution:
y = x + tan x implies that y’ = 1 + sec² x
That is, y’ cos²x = 1 + cos²x.
Differentiating both sides of the above equation we get
y” cos²x + y’ . 2 cos x (-sin x) = 2 cos x (- sin x)
∴ y” cos² x = 2(y’ – 1) sin x cos x
= 2 sec²x sin x cos x = 2 tan x = 2(y – x)
This proves the result.

Question 39.
If x = a(t – sin t),y = a(1 + cost), find \(\frac{d^{2} y}{d x^{2}}\).
Solution:

Question 40.
Find the second order derivative of y = tan^-1(\(\frac{2 x}{1-x^{2}}\))
Solution:
Put x = tan θ, Then
y = tan^-1 (\(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\))
= tan^-1 (tan 2θ)
= 2θ = 2 tan^-1x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{1+x^{2}}\) and \(\frac{d^{2} y}{d x^{2}}\) = \(\frac{-4 x}{\left(1+x^{2}\right)^{2}}\).

Question 41.
If y = sin (sin x), show that y” + (tan x) y’ + y cos²x = 0.
Solution:
y = sin (sin x) implies that .
y’ = cos x . cos (sin x) and
y” = -cos²x sin (sin x) – sin x cos (sin x)
= – y cos²x – sin x (\(\frac{y^{\prime}}{\cos x}\))
= -y cos²x – y’ tan x
∴ y” + (tan x)y’ + y cos² x = 0.

Question 42.
If f(x) = e^x(x ∈ R), then show that f(x) = e^x by first principle.
Solution:
From f(x) = e^x we have for h ≠ 0

Question 43.
If f(x) = log x (x > 0), then show that f(x) = \(\frac{1}{x}\) by first principle.
Solution:
Now for h ≠ 0

\(\frac{d}{dx}\) (log x) = \(\frac{1}{x}\)

Question 44.
If 1(x) = a^x (x ∈ R) (a > 0), then show that f'(x) = a^x log a by first principle.
Solution:
For h ≠ 0
\(\frac{f(x+h)-f(x)}{h}\) = \(\frac{a^{x+h}-a^{x}}{h}\) = a^x [latex]\frac{a^{h}-1}{h}[/latex]
We know that \(\frac{a^{h}-1}{h}\) → log a as h → 0
Hence f'(x) = a^x . log a.
\(\frac{d}{d x}\) = (a^x) = a^x log a

Question 45.
If y = Tan^-1 \(\sqrt{\frac{1-x}{1+x}}\) (|x| < 1), we shall find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Substituting x cos u (u ∈ (0, π)) in y, we get


observe that Tan^-1x, \(\sqrt{\frac{1-x}{1+x}}\) and cos u are the functions that stand for f(x), g(x) and h(u) respectively, mentioned in the method.

Question 46.
If y = Tan^-1[latex]\frac{2 x}{1-x^{2}}[/latex] (|x| < 1) then we shall \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Substituting x = tan u

Question 47.
If x = a cos³t, y = a sin³t, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Here \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3a cos²t (-sin t) and
\(\frac{d y}{d t}\) = 3a sin²t. cost.
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\) = -tan t

Question 48.
If y = e^t +cost, x = log t + sin t find \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Solution:
Here \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^t – sin t and \(\frac{d x}{d t}\) = \(\frac{1}{t}\) + cos t
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{t\left(e^{t}-\sin t\right)}{(1+t \cos t)}\)

Question 49.
To find the derivative of f(x) = x^{\(\sin ^{\frac{1}{x}}\)} with respect to g(x) = sin^-1x, we have to compute \(\frac{d f}{d g}\)
Solution:
Now f(x) = x^{\(\sin ^{\frac{1}{x}}\)} implies that
log f(x) = sin^-1x . log x so that

Question 50.
If x³ + y³ – 3axy = 0, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let the given equation define the function.
y = 1(x) that is x³ + (f(x))³ – 3axf(x) = 0
Differentiating both sides of this equation with respect to x, we get
3x² + 3 (f(x))2 f'(x) – [3a. f(x) + 3axf'(x)] = 0
Hence 3x² + 3y² f'(x) – [3ay + 3ax f'(x)] = 0
∴ f'(x) = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{a y-x^{2}}{y^{2}-a x}\)

Question 51.
If 2x² – 3xy + y² + x + 2y – 8 = 0, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Treating y as a function of x and then differentiating with respect to x,
we get 4x – 3y – 3xy’ + 2yy’ + 1 + 2y’ = 0
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y’ = \(\frac{3 y-4 x-1}{2 y-3 x+2}\)

Question 52.
If y = x^x (x > 0), we shall find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Taking logarithms on both the sides of
y = x^x we obtain log y = x log x
Differentiating with respect to x,
We get \(\frac{y^{\prime}}{y}\) = x . \(\frac{1}{x}\) + log x = 1 + log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y’ = y(1 + log x) = x^x (1 + log x)

Question 53.
If y = (tan x)^{sin x} [o < x < \(\frac{\pi}{2}\) ] compute \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Taking logarthms on both sides of
y= (tan x)^{sin x}, we get
log y = sin x . log (tan x)
Differentiating with respect to x, we get
\(\frac{y^{\prime}}{y}\) = \(\frac{\sin x}{\tan x}\) . sec²x + cosx . log (tan x)
= sec x + cos x . log (tan x)
Hence \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (tan x)^{sin x} [sec x + cos x log (tan x)]

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Limits and Continuity Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Limits and Continuity Important Questions

Question 1.
Evaluate
Solution:

Question 2.
Compute
Solution:
Write f(x) = \(\frac{x-2}{x^{3}-8}\) x ≠ 2 so that
f(x) = \(\frac{x-2}{x^{3}-8}\) = \(\frac{1}{x^{2}+2 x+4}\)
Write h(x) = x² + 2x + 4 so that

Question 3.
Find
Solution:

Question 4.
Compute .
Solution:
For x ≠ 0, we know that -1 ≤ sin \(\frac{1}{x}\) ≤ 1
∴ -x² ≤ x² . sin \(\frac{1}{x}\) ≤ x²

Question 5.
Find .
Solution:
We define f : R → R by f(x) = x² – 5 and g : R → R by g(x) = 4x + 10.

Question 6.
Find .
Solution:
Write F(x) = x³ – 6x² + 9x
= x(x – 3)² = (x – 3)
f(x) where f(x) = x(x – 3)
Write G(x) = x² – 9 = (x – 3) (x + 3)
= (x – 3) g(x) where g(x) = x + 3
∴ \(\frac{F(x)}{G(x)}=\frac{(x-3) f(x)}{(x-3) g(x)}=\frac{f(x)}{g(x)}\)
and g(3) = 6 ≠ 0.
If F and G are polynomials such that f(x) = (x – a)^k, G(x) = (x – a)^k g(x) for some k ∈ N and for some polynomials f(x) and g(x) with

Question 7.
Find .
Solution:
We write F(x) = x³ – 3x² = x²(x – 3) = (x – 3)
f(x) where f(x) = x²,
and G(x) = x² – 5x + 6 = (x – 3)(x – 2)
= (x – 3) g(x) where g(x) = x – 2,
with g(3) = 3 – 2 = 1 ≠ 0.
∴ by applying Theorem g(a) ≠ 0

Question 8.
Show that and (x ≠ 0).
Solution:

Question 9.
Let f : R → R be defined by
f(x) = \(\begin{cases}2 x-1 & \text { if } x<3 \\ 5 & \text { if } x \geq 3\end{cases}\) show that .
Solution:

Question 10.
Show that .
Solution:
Observe that \(\sqrt{x^{2}-4}\) is not defined over (-2, 2)

Question 11.

Solution:

Question 12.
Show that .
Solution:

Question 13.
Find
Solution:
For 0 < |x| < 1, we have

Question 14.
Compute
Solution:
For 0 < |x| < 1

Question 15.
Show that
Solution:

Question 16.
Compute [Mar 13] (a > 0, b > 0, b ≠ 1).
Solution:

Question 17.
Compute , b ≠ 0, a ≠ b.
Solution:

Question 18.
Compute
Solution:

Question 19.
Compute
Solution:

Question 20.
Evaluate
Solution:

Question 21.
Show that
Solution:
Given ε > 0, choose ∞ = \(\frac{1}{\sqrt{\varepsilon}}\) > 0
x > ∞ ⇒ x > \(\frac{1}{\sqrt{\varepsilon}}\) ⇒ x² > \(\frac{1}{\varepsilon}\) ⇒ \(\frac{1}{x^{2}}\) < ε

Question 22.
Show that
Solution:
Given k > 0, let ∞ = log k.
x > ∞ ⇒ e^x ⇒ e^∞ = k

Question 23.
Compute
Solution:

Question 24.
Evaluate .
Solution:

Question 25.
If f(x) \(\frac{a_{n} x^{n}+\ldots+a_{1} x+a_{0}}{b_{m} x^{m}+\ldots+b_{1} x+b_{0}}\) when a_n > 0, b_m > 0, then show that f(x) = ∞ if n > m.
Solution:

Question 26.
Compute
Solution:
-1 ≤ sinx ≤ 1 ⇒ -1 ≤ -sinx ≤ 1
x² – 1 ≤ x² – sinx ≤ x² + 1
Since x → ∞, suppose that the x² – 2 > 0

Question 27.
Show that f(x) = [x] (x ∈ R is continuous at only those real numbers that are not integers.
Solution:
Case i) : If a ∈ z, f(a) = (a) = a

∴ f is not continuous at x = a ∈ z.

Case ii) : If a ∉ z, then ∃ n ∈ z such that n < a < n + 1 then f(a) = (a) = n.

Question 28.
If f : R → R is such that f(x + y) = f(x) + f(y) for all x, y ∈ R then f is continuous on R if it is continuous at a single point in R.
Solution:
Let f be continuous at x₀ ∈ R

∴ f is continuous at x.
Since x ∈ R is arbitrary, f is continuous on R.

Question 29.
Check the continuity of the function f given below at 1 and 2.
f(x) = \(\left\{\begin{array}{cl}
x+1 & \text { if } x \leq 1 \\
2 x & \text { if } 1<x<2 \\
1+x^{2} & \text { if } x \geq 2
\end{array}\right.\)
Solution:

∴ f is continuous at x = 1

f is not continuous at x = 2.

Question 30.
Show that the function f defined on R by f(x) = Cos x², x ∈ R is continuous function.
Solution:
We define h : R → R by h(x) = x² and
g : R → R by g(x) = cosx.
Now, for x ∈ R
have (goh)(x) = g(h(x)) = g(x²)
= cos x² = f(x)
Since g and h continuous on their respective domains, by Theorem
Let A, B, ⊆ R.
Let f : A → R be continuous on A and let
g : B → R be continuous on B.
If f(A) ⊆ B then the composite function
gof : A → R is continous on A.
It follows that a continuous function on R.

Question 31.
Show that the function f defined on R by f(x) = |1 + 2x + |x||, x ∈ R is a continuous function.
Solution:
We define g : R → R by
g(x) = 1 + 2x + |x|, x ∈ R,
and h : R → R by h(x) = |x|, x ∈ R. Then
(hog) (x) = h(g(x)) = h(1 + 2x + |x|)
= |1 + 2x + |x|| = f(x).

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B The Plane Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B The Plane Important Questions

Question 1.
Find the equation of the plane if the foot of the perpendicular from origin to the plane is (2, 3, -5).
Solution:
O (0, 0, 0), P (2, 3, -5) are the given points D.Rs of OP are 2, 3, -5
OP is normal to the plane and it passes through p (2, 3,-5)

The equation of the plane is
2(x – 2) + 3(y – 3) – 5(z + 5)
2x – 4 + 3y – 9 – 5z – 25 = 0
2x + 3y – 5z – 38 = 0

Question 2.
Find the equation to the plane through the points (0, -1, -1) (4, 5, 1) and (3, 9, 4).
Solution:
A (0, -1, -1), B (4, 5, 1), C (3, 9, 4) are the given points.
Equation of the plane through A can be take as
a (x – 0) + b(y + 1) + c (z + 1) = 0 ………………… (1)
The plane passes through B (4, 5, 1), C (3, 9, 4)
4a + 6b + 2c = 0 ………………. (2)
3a + 10b + 5c = 0 ……………… (3)

Substituting in (1) equation of the plane ABC is
5x – 7 (y + 1) + 11 (z + 1) = 0
5x – 7y – 7 + 11z + 11 = 0
5x – 7y + 11z + 4 = 0

Question 3.
Find the equation to the plane parallel to the ZX plane and passing through (0, 4, 4).
Solution:
Equation of ZX plane is y = 0
Equation of the parallel plane is y = k
This plane passes through P(0, 4, 4) = 4 = K
The equation the required plane is y = 4.

Question 4.
Find the equation of the plane through the point, (α, β, γ) and parallel to the plane ax + by + cz = 0.
Solution:
Equation of the given plane is ax + by + cz = 0
Equation of the parallel plane is ax + by+ cz = K
This plane passes through P(α, β, γ)
aα + bβ + cγ = K
∴ Equation of the required plane is
ax + by + cz = aα + bβ + cγ
i.e., a(x – α) + b(y – β) + c(z – γ) = 0

Question 5.
Find the angle between the plane 2x – y + z = 6 and x + y + 2z = 7. [Mar 11]
Solution:
Equation of the planes are
2x – y + z = 6
x + y + 2z = 7
If θ is the angle between the planes, then
cos θ = \(\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\)
= \(\frac{|2.1+(-1) .1+1.2|}{\sqrt{4+1+1} \sqrt{1+1+4}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
= cos π/3
∴ θ = π/3

Question 6.
Find the equation of the plane passing through (2, 0, 1) and (3, -3, 4) and perpendicular to x – 2y + z = 6.
Solution:
Equation of the plane passing through (2, 0, 1) can be taken as
a(x – 2) + by + c(z – 1) = 0 ……………… (1)
This plane passes through B(3, -3, 4)
a – 3b + 3c = 0 ……………… (2)
The plane (1) is perpendicular to x- 2y + z = 6
a – 2b + c = 0 ………………… (3)

Substituting in (1), equation of the required plane is
3(x – 2) + 2y + 1 (z – 1) = 0
3x – 6 + 2y + z – 1 = 0
3x + 2y + z – 7 = 0
3x + 2y + z = 7.

Question 7.
Let us reduce the equation of the plane x + 2y – 2z – 9 = 0 to the normal form and hence find the direction cosines of the normal to the plane and the length of the perpendicular drawn from the origin to the given plane.
Solution:
The equation of the given plane is
x + 2y – 2z – 9 = 0
Bringing the constant term to R.H.S.
x + 2y – 2z = 9 ………………. (1)
Square root of the sum of the squares of the coefficients of x, y, z in (1) is
\(\sqrt{1^{2}+2^{2}+2^{2}}\) = ±3
dividing (1) by 3, we observe that
p = ± (\(\frac{-9}{3}\)) = ±3
±\(\frac{1}{3}\)x ±\(\frac{2}{y}\)y ±\(\frac{2}{3}\)z = ±3
Choosing the sign of the equation so that the constant on the right is positive, we get,
\(\frac{x}{3}\) + \(\frac{2}{3}\)y – \(\frac{2}{3}\)z = 3 ………………. (2)
(2) represents the equation of the plane in the normal form. Hence d.c.’s of the normal to the plane are (\(\frac{1}{2}\), \(\frac{2}{3}\), –\(\frac{2}{3}\)) and the length of the perpendicular from the origin to the plane is 3.

Question 8.
Suppose a plane makes intercepts 2, 3, 4 on the X, Y, Z – axes respectively. The equation of the plane whose X, Y, Z intercepts are a, b, c is \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1.
Solution:
From the equation of the plane is
\(\frac{x}{2}\) + \(\frac{y}{3}\) + \(\frac{z}{4}\) = 1.
i.e., 6x + 4y + 3z = 12

Question 9.
Consider the plane whose equation is x – 3y + 2z = 9.
Solution:
Dividing by 9, \(\frac{x}{9}\) + \(\frac{y}{-3}\) + \(\frac{z}{9 / 2}\) = 1
Comparing this with \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
a = X-intercept = 9
b = Y-intercept = 3
c = Z-intercept = \(\frac{9}{2}\)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions

Question 1.
If P (2, 3, -6), Q (3, -4, 5) are two points, find the d.c.s of \(\overrightarrow{\mathrm{OP}}\), \(\overrightarrow{\mathrm{QO}}\) and \(\overrightarrow{\mathrm{PQ}}\) where O is the origin.
Solution:

Question 2.
Find the d.c.s of a line that makes equal angles with the axes.
Solution:
Suppose the line makes angle a with the axes d.cs of the line are (cos α, cos α, cos α)
But l² + m² + n² = 1
cos² α + cos² α + cos² α = 1
3 cos² α = 1 ⇒ cos α = \(\frac{1}{3}\)
cos α = ± \(\frac{1}{\sqrt{3}}\)
d.cs of the required line are
(± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\))
Hence there are 8 such directions which reduce to 4 lines.

Question 3.
If the d.cs of a line are (\(\frac{1}{c}\), \(\frac{1}{c}\), \(\frac{1}{c}\)) find c.
Solution:
We know that l² + m² + n² = 1
\(\frac{1}{c^{2}}\) + \(\frac{1}{c^{2}}\) + \(\frac{1}{c^{2}}\) = 1, \(\frac{3}{c^{2}}\) = 1 ⇒ c² = 3
c = ± \(\sqrt{3}\)

Question 4.
Find the direction cosines of two lines which are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0. [Mar 11]
Solution:
Given l + m + n = 0 ………………… (1)
mn – 2nl – 2lm = 0 ………………….. (2)
From (1)l = -(m + n)
Substituting in (2)
mn ± 2n (m + n) + 2m (m + n) = 0
mn + 2mn + 2n² + 2m² + 2mn = 0
2m² + 5mn + 2n² = 0
(2m + n) (m + 2n) = 0
2m = -n or m = -2n
Case (i): 2m₁ = -n₁
From l_r = -m₁ – n₁
= -m₁ + 2m₁ = m₁
\(\frac{l_{1}}{1}\) = \(\frac{m_{1}}{1}\) = \(\frac{x_{1}}{-2}\)
D.Rs of the first line are 1, 1, -2
D.Cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\), –\(\frac{2}{\sqrt{6}}\)
Case (ii): m₂ = -2n₂
From (l) l₂ = -m₂ – n₂ = +2n₂ – n₂ = n₂
\(\frac{l_{2}}{1}\) = \(\frac{m_{2}}{-2}\) = \(\frac{n_{2}}{1}\)
d.cs of the second line are 1, -2, 1
d.cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{-2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\)

Question 5.
A ray makes angles \(\frac{\pi}{3}\), \(\frac{\pi}{3}\) with \(\overrightarrow{\mathrm{OX}}\) and \(\overrightarrow{\mathrm{OY}}\) respectively. Find the angle made by it with \(\overrightarrow{\mathrm{OZ}}\).
Solution:
We know that l² + m² + n² = 1
cos²α + cos²β + cos²γ = 1
i.e., cos² \(\frac{\pi}{3}\) + cos² \(\frac{\pi}{3}\) + cos² γ = 1
\(\frac{1}{4}\) + \(\frac{1}{4}\) + cos² γ = 1
cos² γ = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
cos γ = ± \(\frac{1}{\sqrt{2}}\)
γ = cos^-1 (± \(\frac{1}{\sqrt{2}}\))
= \(\frac{\pi}{4}\) or \(\frac{3\pi}{4}\).

Question 6.
Find the d.r.s and d.c.s of the line joining the points (4, -7, 3), (6, -5, 2).
Solution:
d.c.s of the lines are (6 – 4, -5 + 7, 2 – 3)
= (2, 2, -1)
Dividing with \(\sqrt{4+4+1}\) = 3
d.cs of the line are ± (\(\frac{2}{3}\), \(\frac{2}{3}\), –\(\frac{1}{3}\)

Question 7.
If the d.c.s of a line are proportional the (1, -2, 1) find its d.c.s.
Solution:
d.rs of the line are (1, -2, 1 )
Dividing with \(\sqrt{6}\)
d.cs of the line are ± (\(\frac{1}{\sqrt{6}}\), –\(\frac{2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\))

Question 8.
Show that the line joining the points P(0, 1, 2) and Q (3, 4, 8) is parallel to the line joining the points R (-2. \(\frac{3}{2}\). – 3) and S (\(\frac{5}{2}\), 6, 6).
Solution:
DR’s of PQ are (3 – 0, 4 – 1, 8 – 2) = (3, 3, 6)
DR’s of RS are (\(\frac{5}{2}\) + 2, 6 – \(\frac{3}{2}\), 6 + 3)
= (\(\frac{9}{2}\), \(\frac{9}{2}\), 9)
d.r’s of PQ and RS are proportional since \(\frac{2}{3}\) (\(\frac{9}{2}\), \(\frac{9}{2}\), 9) = (3, 3, 6)
∴ PQ is parallel to RS.

Question 9.
Show that the line joining the points A (2, 3, -1) and B(3, 5, -3) is perpendicular to the Line joining C(1, 2, 3) and D(3, 5, 7).
Solution:
d.rs of AB are (3 – 2, 5 – 3, -3 + 1)
= (1, 2, -2)
d.rs of CB are (3 – 1, 5 – 2, 7 – 3) = (2, 3, 4)
a₁a₂ + b₁b₂ + c₁c₂ = 1.2 + 2.3 – 2.4
= 2 + 6 – 8 = 0
∴ AB and CD are perpendicular.

Question 10.
For what value of x the line joining A(4, 1, 2) B (5, x, 0) is perpendicular to the line joining C(1, 2, 3) and D(3, 5, 7) ?
Solution:
d.r’s of AD are (1, x-1, -2)
d.rs of CD are (2, 3, 4)
AB and CD are perpendicular
a₁a₂ + b₁b₂ + c₁c₂ = 0
1.2 – 3(x – 1) + 4 (-2) = 0
2 + 3x – 3 – 8 = 0
3x = 9
⇒ x = 3

Question 11.
Show that the points A (1, 2, 3), B (4, 0, 4), C(-2, 4, 2) are collinear.
Solution:
d.rs of \(\overline{\mathrm{AB}}\) are 4 – 1, 0 – 2, 3 – 4
i.e., 3, -2, -1
d.rs of \(\overline{\mathrm{BC}}\) are -2 – 4, 4 – 0, 2 – 4
i.e., -6, 4, -2
d.rs of AB and BC are proportional and B is a common point. A, B, C are collinear.

Question 12.
A (1, 8, 4), B (0, -11, 4), C (2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Solution:
Suppose divides BC in the ratio m = n
Co-ordinates of D are

(\(\frac{2 m-m-n}{m+n}\), \(\frac{-3 m-11 n-8 m-8 n}{m+n}\), \(\frac{m+4 n-4 m-4 n}{m+n}\))
\(\frac{m-n}{m+n}\), \(\frac{-11 m-19 n}{m+n}\), \(\frac{-3 m}{m+n}\)
d.rs of BC are (2 – 0, -3 + 11, 1 – 4) = (2, 8, -3)
AB and BC are perpendicular
\(\frac{2(m-n)}{m+n}\) + 8\(\frac{(-11 m-19 n)}{m+n}\) + \(\frac{(-3)(-3 m)}{m+n}\) = 0
2m – 2n – 88m – 152n + 9m = 0
-77m – 154n = 0
77m = -154n
m = -2n
co-ordinates of are
(\(\frac{-4 n}{-n}\), \(\frac{6 n-11 n}{-n}\), \(\frac{-2 n+4 n}{-n}\))
= (4, 5, -2)

Question 13.
Lines \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) are drawn from O with direction cosines proportional to (1, -2, -1), (3, -2, 3). Find the direction cosines of the normal to the plane AOB.
Solution:
Suppose l, m, n are the d.cs of the normal.
It is perpendicular to every line in the plane AOB
l – 2m – n = 0
3l – 2m + 3n = 0

Question 14.
Find the angle between the diagonals of a cube.

Solution:
Let ‘O’ one of the vertices of the cube taken as origin and the three sides OA, OB, OC are taken as co-ordinate axes. Let OA = OB = OC = a the four diagonals are \(\overrightarrow{\mathrm{OF}}\), \(\overrightarrow{\mathrm{AG}}\), \(\overrightarrow{\mathrm{DE}}\) and \(\overrightarrow{\mathrm{BC}}\)
The co-ordinates of the vertices of the cube are 0(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0,0, a), F(a, a, 0), D(a, a, 0), E(a, 0, a), G(0, a, a)
D.Rs of OF are (a-0, a – 0, a – 0) = (a, a, a)
D.Rs of AG are (0 – 9, a – 0, a – 0) = (-a, a, a)
If O is the angle between the diagonals OF and AG then
cos θ = \(\frac{|a(-a)+a \cdot a+a \cdot a|}{\sqrt{a^{2}+a^{2}+a^{2}} \cdot \sqrt{a^{2}+a^{2}+a^{2}}}\)
= \(\frac{a^{2}}{3 a^{2}}\) = \(\frac{1}{3}\)
θ = Cos^-1 (\(\frac{1}{3}\))y
Similarly, the angle between any pair of diagonals can be shown to be cos^-1 (\(\frac{1}{3}\))

Question 15.
Show that the line whose d.c’s are proportional to (2, 1, 1) (4, \(\sqrt{3}\) -1, –\(\sqrt{3}\) – 1) are inclined to one another at angle \(\frac{\pi}{3}\).
Solution:
The d.cs of the given lines are (2, 1, 1), (4, \(\sqrt{3}\) – 1, –\(\sqrt{3}\) – 1) 4² + (\(\sqrt{3}\) – 1)² + (-\(\sqrt{3}\) – 1)²
= 16 + 3 + 1 – 2\(\sqrt{3}\) + 3 + 1 + 2\(\sqrt{3}\)
= 24

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Three Dimensional Coordinates Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Three Dimensional Coordinates Important Questions

Question 1.
Show that the points A(-4, 9, 6), B(-1, 6, 6), and C(0, 7, 10) from a right-angled isosceles triangle.
Solution:
A(-4, 9, 6), B(-1, 6, 6), C(0, 7, 10) are the vertices of ∆ABC.
AB = \(\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2}}\)
= \(\sqrt{9+9}\)
= \(\sqrt{18}\)
BC = \(\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}}\)
= \(\sqrt{1+1+16}\) = \(\sqrt{18}\)
CA = \(\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}\)
= \(\sqrt{16+4+16}\) = \(36\)
AB = BC and AB² + BC² = CA²
ABC is a right-angled isosceles triangle.

Question 2.
Show that locus of the point whose distance from Y-axis is thrice its distance from (1, 2, -1) is 8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0.
Solution:
Let P (x, y, z) be any point on the locus PM – distance from Y – axis = \(\sqrt{x^{2}+z^{2}}\)
A(1, 2, -1) is the given point
Given condition is PM = 3 . PA
PM² = 9PA²
x² + z² = 9[(x – 1)² + (y – 2)² + (z + 1)²]
= 9x² – 18x + 9 + 9y² – 36y + 36 + 9z² + 18z + 9
Locus of P is 8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0
P satisfies the equation
8x² + 9y² + 8z² – 18x – 36y + 18z + 54 = 0

Question 3.
A, B, C are three points on \(\overrightarrow{o x}\), \(\overrightarrow{o y}\), \(\overrightarrow{o z}\) respectively at distances a, b, e. (a ≠ 0, b ≠ 0, c ≠ 0) from the origin 0. Find the co-ordinates of the point which is equidistant from A, B, C and O.
Solution:
A is a point on ox ⇒ co-ordinates of A are (a, 0, 0)
Similarly co-ordinates of B are (0, b, 0) and co-ordinates of C are (0, 0, c)
P(x, y, z) is the required point
PO = PA = PB = PC
PO² = PA² = PB² = PC²
PO² = PA²
x² + y² + z² = (x – a)² + y² + z²
x² – x² + a² – 2ax = 0
2ax = a² ⇒ x = \(\frac{a^{2}}{2 a}\) = \(\frac{a}{2}\)
PO² = PB² ⇒ y = b/2
PO² = PC² ⇒ z = c/2
Co-ordinates of P are (\(\frac{a}{2}\), \(\frac{b}{2}\), \(\frac{c}{2}\) )

Question 4.
Show that the points A (3, -2, 4) B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
A(3, -2, 4), B(1, 1, 1), C(-1, 4, -2) are the given points

A, B, C are collinear.

Question 5.
Find the ratio in which YZ-plane divides the line joining A(2, 4, 5) and B(3, 5, -4). Also find the point of intersection
Solution:
Suppose the line AB meets YZ plane in P
Let P divide AB in the ratio k : 1
Co-ordinates of P are
\(\left(\frac{3 k+2}{k+1}, \frac{5 k+4}{k+1}, \frac{-4 k+5}{k+1}\right)\)
P is a point on YZ plane
⇒ x co-ordinates of p = 0
\(\frac{3 k+2}{k+1}\) = 0 ⇒ 3k + 2 = 0
k = –\(\frac{2}{3}\)
YZ plane divides AB is the ratio -2 : 3
Substituting the value of k in co-ordinates of p, co-ordinates of p are
\(\left[0, \frac{5\left(-\frac{2}{3}\right)+4}{-\frac{2}{3}+1} \cdot \frac{(-4)\left(-\frac{2}{3}\right)+5}{-\frac{2}{3}+1}\right]\)
(0, 2, 23)

Question 6.
Show that the points A(3, -2, 4), B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
Suppose the point P divides AD in the ratio k : 1
Co-ordinates of P are \(\left(\frac{k+3}{k+1}, \frac{k-2}{k+1}, \frac{k+4}{k+1}\right)\)
If A, B, C are collinear, C lies on AB
P must coincide with C for same value of k
\(\frac{k+3}{k+1}\) = -1 ⇒ k + 3 = -k – 1
2k = 4 ⇒ k = -2
Substituting k = -2 in co—ordinates of P we get
\(\left(\frac{-2+3}{-2+1}, \frac{-2-2}{-2+1}, \frac{-2+4}{-2+1}\right)\)
= (-14, -2) = c
A, B, C are collinear.

Question 7.
Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) are (4, 5, 1) [Mar 11]
Solution:

ABCD is a parallelogram
where A = (2, 4, -1), B = (3, 6, -1) C = (4, 5, 1)
Suppose D(x, y, z) is the fourth vertex
A B C D is a parallellelogram
Mid point of AC = Mid point of BD

∴ Co-ordinates of the fourth vertex are = D (3, 3, 1)

Question 8.
A(5, 4, 6), B(1, -1, 3), C(4, 3, 2) are three points. Find the co-ordinates of the point in which the bisector of ∠BAC meets the side \(\overline{B C}\).
Solution:
We know that if AB is the bisector of ∠BAC
Divides BC in the ratio AB : AC

Question 9.
If (x₁, y₁, z₁) and (x₂, y₂, z₂) are two vertices and (α, ß, γ) is the centroid of a triangle, find the third vertex of the triangle.
Solution:
Let A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂) be the two vertices of the triangle ABC.

Let G = (α, β, γ) be the centroid.
If C = (x₃, y₃, z₃) is the third vertex, then we have
(\(\frac{x_{1}+x_{2}+x_{3}}{3}\), \(\frac{y_{1}+y_{2}+y_{3}}{3}\), \(\frac{z_{1}+z_{2}+z_{3}}{3}\)) = (α, β, γ)
⇒ x₁ + x₂ + x₃ = 3α; y₁ + y₂ + y₃ = 3β; z₁ + z₂ + z₃ = 3γ.
⇒ x₃ = 3α – x₁ – x₂; y₃ = 3β – y₁ – y₂; z₃ = 3γ – z₁ – z₂.
∴ The third vertex
C = (3α – x₁ – x₂, 3β – y₁ – y₂, 3γ – z₁ – z₂).

Question 10.
If D(x₁, y₁, z₁), E(x₂, y₂, z₂) and F(x₃, y₃, z₃) are the midpoints of the sides BC, CA and AB respectively of a triangle, find its vertices A, B and C.
Solution:
It is given that D is the mid point of the side BC, E is the mid point of the side CA and F is the mid point of the side AB.

∴ DEP is the triangle formed out of the mid points of the three sides.
Consider the parallelogram AEDF. Let A = (h, k, s).
Mid point of AD = Mid point of EF
⇒ (\(\frac{h+x_{1}}{2}\), \(\frac{k+y_{1}}{2}\), \(\frac{s+z_{1}}{2}\)) = (\(\frac{x_{2}+x_{3}}{2}\), \(\frac{y_{2}+y_{3}}{2}\), \(\frac{z_{2}+z_{3}}{2}\))
⇒ h = x₂ + x₃ – x₁; k = y₂ + y₃ – y₁; s = z₂ + z₃ – z₁
∴ Vertex A
= (x₂ + x₃ – x₁, y₂ + y₃ – y₁, z₂ + z₃ – z₁)
Similarly, the vertices B and C can be obtained as
B = (x₃ + x₁ – x₂, y₃ + y₁ – y₂, z₃ + z₁ – z₂)
C = (x₁ + x₂ – x₃, y₁ + y₂ – y₃, z₁ + z₂ – z₃).

Question 11.
If M(α, β, γ) is the mid point of the line segment joining the points A(x₁, y₁, z₁) and B, then find B.
Solution:

Let B(h, k, s) be the point required.
It is given that M is the mid point of AB.
∴ We have (α, β, γ) = (\(\frac{x_{1}+h}{2}\), \(\frac{y_{1}+k}{2}\), \(\frac{z_{1}+s}{2}\))
⇒ 2α = x₁ + h; 2β = y₁ + k; 2γ = z₁ + s
⇒ h = 2α – x₁; k = 2β – y₁; s = 2γ – z₁
∴ Point Bis (2α – x₁, 2β – y₁, 2γ – z₁).

Question 12.
If H, G, S and I respectively denote orthocentre, centroid, circumcentre and in-centre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.
Solution:

Since AB = BC = CA, ABC is equilateral triangle.
We know that orthocentre, centroid, circumcentre and incentre of equilateral triangle are the same (i.e., all the four points coincide). Now, centroid .
G = (\(\frac{1+2+3}{3}\), \(\frac{2+3+1}{3}\), \(\frac{3+1+2}{3}\))
= (2, 2 ,2)
∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2).

Question 13.
Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).
Solution:
If a, b, care the side of the triangle ABC, where
A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃) are the vertices, then the in-centre of the triangle is given by

Question 14.
If the point (1, 2, 3) is changed to the point (2, 3, 1) through translation of axes, find the new origin.
Solution:
Let (x, y. z) be the co-ordinates of any point P w.r.t. the co-ordinate frame Oxyz and (X, Y, Z) be the co-ordinates of P w.r.t the new frame of reference O’XYZ.

Let O’ (h, k, s) be the new origin so that
x = X + h, y = Y + k and z = Z + s.
⇒ (h, k, s) (x – X, y – Y, z – Z)
⇒ (h, k, s) = (1 – 2, 2 – 3, 3 – 1)
= (-1, -1, 2).
∴ O’ = (-1, -1, 2) is the new origin.

Question 15.
Find the ratio in which the point P(5, 4, -6) divides the line segment joining the points (3, 2, -4) and B(9, 8, -10). Also find the harmonic conjugate of P.
Solution:

∴ Q(-3, -4, 2) is the harmonic conjugate of P(5, 4, -6).

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Pair of Straight Lines Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Pair of Straight Lines Important Questions

Question 1.
Does the equation x² + xy + y² = 0 represent a pair of lines?
Solution:
a = 1, b = 1, h = \(\frac{1}{2}\), ⇒ h² = \(\frac{1}{4}\), ab = 1
h² = ab < 0 i.e., h² < ab.
∴ The given equation does not represent a pair of lines

Question 2.
Find the nature of the triangle of formed by the lines x² – 3y² = 0 and x = 2.
Solution:
Combined equation of OA and OB is
x² – 3y² = 0
(x + \(\sqrt{3}\) y) (x – \(\sqrt{3}\) y) = 0
x + \(\sqrt{3}\) y = 0 and x – \(\sqrt{3}\) y = 0
i.e., y = \(\frac{1}{\sqrt{3}}\) x, y = –\(\frac{1}{\sqrt{3}}\) x are equally inclined to the X—axis, the inclination being 30°
∴ ∠OAB – ∠OBA = 60°
∴ The triangle is equilateral

Question 3.
Find the centroid of the triangle formed by the lines 12x² – 20xy + 7y² = 0 and 2x – 3y + 4 = 0.
Solution:
Combined equation of OA and OB is
12x² – 20xy + 7y² = 0 ………………….. (1)
Equation of AB is 2x – 3y + 4 = 0
2x = 3y – 4
Substituted in (1)
3(3y – 4)² – 1oy (3y – 4) + 7y² = 0
3(9y² + 16 – 24y) – 30y² + 40y + 7y² = 0
27y² +48 – 72y – 30y² + 40y + 7y² = 0
4y² – 32y + 48 = 0
y² – 8y + 12 = 0
(y – 2) (y – 6) = 0 ⇒ y = 2 or 6
x = \(\frac{3 y-4}{2}\)
y = 2 ⇒ x = \(\frac{6-4}{2}\) = 1
y = 6 ⇒ x = \(\frac{18-4}{2}\) = 7
7 2
∴ Vertices and O (0, 0), A (1, 2), B( 7, 6)
Centroid of OAB is
(\(\frac{0+1+7}{3}\), \(\frac{0+2+6}{3}\)) = (\(\frac{8}{3}\), \(\frac{8}{3}\))

Question 4.
Prove that the lines represented by the equations x² – 4xy + y² = 0 and x + y = 3 form an equilateral triangle.
Solution:
Since the straight line L: x + y = 3 makes 45° with the negative direction of the X-axis, none of the lines which makes 60° with the line L is vertical. If ‘m’ is the slope of one such straight line, then

\(\sqrt{3}\) = tan 60° = \(\left|\frac{m+1}{1-m}\right|\) and so, m satisfies the equation (m + 1)² = 3(m – 1)²
(or) m² – 4m + 1 = 0 …………………….. (1)
But the straight line having slope’m1 and passing through the origin is
y = mx ……………………. (2)
So the equation of the pair of lines passing through the origin and inclined at 60° with the line L is obtained by eliminating’m’ from the equations (1) and (2). Therefore the combined equation of this pair of lines is
(\(\frac{y}{x}\))² – 4(\(\frac{y}{x}\)) + 1 = 0 (i.e.,) x² – 4xy + y² = 0
Which is the same as the given pair of lines. Hence, the given triad of lines form an equilateral triangle.

Question 5.
Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\). [May 11, 07; Mar. 07, 04]
Solution:
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y)
Then the separate equations of the lines represented by the equation
ax² + 2hxy + by² = 0 are
L₁ : l₁x + m₁y = 0 and L₂ : l₁x + m₁y = 0
Also, we have l₁l₂ = a; m₁m₂ = b and
l₁m₂ + l₂m₁ = 2h
d₁ = length of the perpendicular from (α, β) to L₁ = \(\frac{\left|l_{1} \alpha+\mathrm{m}_{1} \beta\right|}{\sqrt{l_{1}^{2}+\mathrm{m}_{1}^{2}}}\)
d₂ = length of the perpendicular from (α, β) to L₂ = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)
Then, the product of the lengths of the perpendiculars from (α, β) to the given pair of lines = d₁d₂
= \(\frac{\left\|\left(l_{1} \alpha+m_{1} \beta\right)\left(l_{2} \alpha+m_{2} \beta\right)\right\|}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)
= \(=\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Question 6.
Let ax² + 2hxy + by² = 0 represent a pair of straight lines. Then show that the equation of the pair of straight lines.
i) Passing through (x₀, y₀) and parallel to the given pair of lines is
a(x – x₀)² + 2h(x – x₀) (y – y₀) + b(y – y₀)² = 0 and
ii) Passing through (x0, y0) and perpendicular to the given pair of lines is
b(x – x₀)² – 2h(x – x₀) (y – y₀) + a(y – y₀)² = 0
Solution:
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y).
Then the lines L₁ and L₂ are represented by the given equation are respectively
l₁x + m₁y = 0 and l₂x + m₂y = 0.
Also we have l₁l₂ = a, m₁m₂ = b and
l₁m₂ + l₂m₁ = 2h

(i) The straight lines passing through (x₀, y₀) and parallel to L₁ and L₂ are respectively
l₁x + m₁y = l₁x₀ + m₁y₀ (or)
l₁(x – x₀) + m₁(y – y₀) = 0 and
l₂(x -x₀) + m₂(y – y₀) = 0.
Therefore, their combined equation is
[l₁(x – x₀) + m₁(y – y₀)] [l₂(x – x₀)+ m₂(y – y₀)] = 0
(or) a(x – x₀)² + 2h(x – x₀) (y – y₀) + b(y – y₀)² = 0

(ii) The straight lines passing through (x₀, y₀) and perpendicular to the pair L₁ and L₂ are respectively.
m₁x – l₁y = m₁x₀ – l₁y₀ (or)
m₁(x – x₀) – l₁(y – y₀) = 0 and
m₂(x – x₀) – l₂(y – y₀) = 0.
Hence their combined equation is
[m₁(x – x₀) – l₁(y – y₀)] [m₂(x – x₀) – l₂(y – y₀)] = 0
(i.e.,) b(x – x₀)² – 2h(x – x₀) (y – y₀) + a(y – y₀)² = 0
[Note : The pair of lines passing through the origin and perpendicular to the pair of lines given by ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0].

Question 7.
Show that the area of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\left|\frac{n^{2} \sqrt{h^{2}-a b}}{a m^{2}-2 h / m+b l^{2}}\right|\)
Solution:
Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation
ax² + 2hxy + by² = 0 (see figure) and \(\overleftrightarrow{\mathrm{AB}}\) be the line lx + my + n = 0
Let ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y) and\(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the lines.
l₁x + m₁y = 0 and l₂x + m₂y = 0 respectively.

Let A = (x₁, y₁) and B = (x₂, y₂).
Then l₁x₁ + m₁y₁ = 0 and l₁x₁ + m₁y₁ + n = 0.
So, by the rule of cross-multiplication, we obtain

Question 8.
Two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1,0). Find the equation of the third side.
Solution:
Let the lines 7x – y + 3 – 0 and x + y – 3 = 0 intersect at A. If we draw lines (not passing through A) perpendicular to each of the bisectors of the angles at A, we get isosceles triangles, equal sides being along the given lines.
(∆ABF ≅ ∆AFC and ∆ADG ≅ ∆AGE)
Of them, we require those triangles whose third sides pass through (1, 0).
The equations of the bisectors of the angles between 7x – y + 3 = 0 and x + y – 3 = 0 are

⇒ 7x – y + 3 = ± 5(x + y – 3)
⇒ x – 3y + 9 = 0 and 3x + y – 3 = 0.
The third sides will be those lines perpendicular to the bisectors and intersecting at (1, 0).

The side perpendicular to x – 3y + 9 = 0 and passing through (1, 0) is 3x + y – 3 =0. The other one is (x – 1) -3(y – 0) = 0 i.e., x – 3y – 1 = 0. Therefore 3x + y- 3 = 0 and x- 3y— 1 = 0 are the required ones. [In the Figure ∆ABC and ∆ADE are isosceles with \(\overline{B C}\) and \(\overline{D E}\) as third sides].

Question 9.
Find the angle between the straight lines represented by 2x² + 5xy + 2y² – 5x – 7y + 3 = 0.
Solution:
a = 2, b = 2, h = \(\frac{5}{2}\)
cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}=\frac{|2+2|}{\sqrt{(2-2)^{2}+4 \cdot \frac{25}{4}}}\) = \(\frac{4}{5}\)
θ = cos^-1 (\(\frac{4}{5}\))

Question 10.
Find the equation of the pair of lines passing through the origin and parallel to the pair of lines 2x² + 3xy – 2y² – 5x + 5y – 3 = 0
Solution:
Equation of the given pair of lines is
2x² + 3xy – 2y² – 5x + 5y – 3 = 0
Equation of the pair of parallel lines passing through the origin is ax² + 2hxy + by² = 0 is 2x² + 3xy – 2y² = 0

Question 11.
Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0
Solution:
Equation to the pair of lines parallel to the given lines and passing through the origin is ax² + 2hxy + by² = 0
Equation of the pair of lines perpendicular to the given lines and passing through the origin is bx² – 2hxy + ay² = 0

Question 12.
If x² + xy – 2y² + 4x – y + k = 0 represents a pair of straight lines, find k.
Solution:
a = 1, b = -2, c = k; f = –\(\frac{1}{2}\), g = 2, h = \(\frac{1}{2}\)
The condition is
abc + 2fgh – af² – bg² – ch² = 0
2k + 2(-\(\frac{1}{2}\)) . 2\(\frac{1}{2}\) – 1.\(\frac{1}{4}\) + 2.\(\frac{4}{a}\) – k\(\frac{1}{4}\) = 0
-8k – 4 – 1 + 8 – k = 0
9k = 27 ⇒ k = 3

Question 13.
Prove that the equation 2x² + xy – 6y² + 7y – 2 = 0 represents a pair of straight lines.
Solution:
a = 2
b = -6
c = -2
f = \(\frac{7}{2}\)
g = 0
h = \(\frac{1}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-6) (-2) + 2.\(\frac{7}{2}\) 0\(\frac{1}{2}\) – 2(\(\frac{7}{2}\))² + 6.0 + 2.(\(\frac{1}{2}\))²
= 24 – \(\frac{49}{2}\) + \(\frac{1}{2}\) = 0
h² – ab = \(\frac{1}{4}\) + 12 > 0,
g² – ac = 0 + 4 = 4 > 0
f² – bc = \(\frac{49}{4}\) – 12 = \(\frac{1}{4}\) > 0
The given equation represents a pair of line.

Question 14.
Prove that the equation 2x² + 3xy – 2y² – x + 3y – 1 = 0 represents a pair of perpendicular straight lines.
Solution:
a = 2
b = -2
c = -1
f = \(\frac{3}{2}\)
g = –\(\frac{1}{2}\)
h = \(\frac{3}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-2) (-1) + 2.\(\frac{3}{2}\) (-\(\frac{1}{2}\)).\(\frac{3}{2}\) – 2(\(\frac{9}{4}\)) + 2.\(\frac{1}{4}\) + \(\frac{1.9}{2}\)
= 4.-\(\frac{9}{4}\) – 2.\(\frac{9}{4}\) + \(\frac{1}{2}\) + \(\frac{9}{4}\)
= \(\frac{9}{2}\) – \(\frac{9}{2}\) = 0
h² – ab = \(\frac{9}{4}\) + 4 = \(\frac{25}{4}\) > 0,
g² – ac = \(\frac{1}{4}\) + 2 > 0
f² – bc = \(\frac{9}{4}\) – 2 = \(\frac{1}{4}\) > 0
a + b = 2 – 2 = 0
The given equation represents a pair of perpendicular line.

Question 15.
Show that the equation 2x² – 13xy – 7y² + x + 23y – 6 = 0 represents a pair of straight lines. Also find the angle between them and the co-ordinates of the point of intersection of the lines.
Solution:
Here a = 2
b = -7
c = -6
f = \(\frac{23}{2}\)
g = \(\frac{1}{2}\)
h = –\(\frac{13}{2}\)
abc + 2fgh – af² – bg² – ch²
= 2(-7) (-6) + 2.\(\frac{23}{2}\) . \(\frac{23}{2}\) . (-\(\frac{13}{2}\)) – 2\(\frac{529}{4}\) + 7.\(\frac{1}{4}\) + 6 . \(\frac{169}{4}\)
= \(\frac{1}{4}\) (336 – 299 – 1058 + 7 + 1014)
= \(\frac{1}{4}\) (1357 – 1357) = 0
h² – ab = \(\frac{169}{4}\) + 14 > 0,
g² – ac = \(\frac{1}{4}\) + 12 > 0,
f² – bc = \(\frac{529}{4}\) – 42 > 0
The given equation represents a pair of lines

= \(\frac{-13-92}{-56-169}\) = \(\frac{-105}{-225}\) = \(\frac{7}{15}\)
Point of intersection is P (\(\frac{19}{15}\), \(\frac{7}{15}\))

Question 16.
Find that value of λ for which the equation λx² – 10xy + 12y² + 5x – 16y -3 = 0 represents a pair of straight lines.
Solution:
Here a = λ
b = 12
c = -3
f = -8
g = \(\frac{5}{2}\)
h = -5
The given equation represents a pair of lines
abc + 2fgh – af² – bg² – ch² = 0
-36λ + 2(-8)\(\frac{5}{2}\) (-5) – λ . 64 – 12 . \(\frac{25}{4}\) + 3 . 25 = 0
-36λ + 200 – 64λ – 75 + 75 = 0
100λ = 200
⇒ λ = 2 ⇒ a = 2
h² – ab = 25 – 24 = 1 > 0
f² – bc = 64 + 36 = 100 > 0
g² – ac = \(\frac{25}{4}\) + 6 = \(\frac{49}{4}\) > 0
∴ The given equation represents a pair of lines for λ = 2.

Question 17.
Show that die pairs of straight lines 6x² – 5xy – 6y² = 0 and 6x² – 5xy – 6y² + x + 5y – 1 = 0 form a square.
Solution:
H ≅ 6x² – 5xy – 6y² = (3x + 2y) (2x – 3y)
and S ≅ 6x² – 5xy – 6y² + x + 5y – 1
= (3x + 2y- 1) (2x – 3y + 1).
∴ H = 0 represents the lines 3x + 2y = 0 and 2x – 3y = 0 which are perpendicular and S = 0 represents the lines 3x + 2y – 1 = 0, 2x – 3y + 1 = 0 which are also perpendicular. These two pairs of perpendicular lines, therefore, determine a rectangle.
Also the distance between the pair of opposite sides 3x + 2y = 0 and 3x + 2y – 1 = 0 is \(\frac{1}{\sqrt{13}}\) and this is the same as the distance between the other pair of opposite sides 2x – 3y = 0 and 2x – 3y + 1 = 0 of the rectangle. Hence the rectangle is a square.

Question 18.
Show that the equation 8x² – 24xy + 18y² – 6x + 9y – 5 = 0 represents a pair of parallel straight lines and find the distance between them.
Solution:
S = 8x² – 24xy + 18y² – 6x + 9y – 5
= 2(2x – 3y)² – 3(2x – 3y) – 5
= [2(2x – 3y) – 5] [(2x – 3y) + 1]
= (4x – 6y – 5) (2x – 3y + 1) = 0
The lines are 4x – 6y – 5 = 0 and 2x – 3y + 1 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\), \(\frac{b_{1}}{b_{2}}\) = \(\frac{-6}{-3}\) = 2
\(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\)
∴ The given equation represents a pair of parallel lines.
Distance between the lines
= 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) = 2\(\sqrt{\frac{9+40}{8(8+18)}}\) = \(\frac{2.7}{2 \sqrt{52}}\) = \(\frac{7}{\sqrt{52}}\)

Question 19.
If the pair of lines represented by ax² + 2hxy + by² = 0 and ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus, prove that (a – b) fg + h(f² – g²) = 0.
Solution:
Combined equation of OA and OB is ax² + 2hxy + by² = 0
Combined equation of AC, BC is ax² + 2hxy + by² + 2gx + 2fy + c = 0
Point of intersection is C(\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g h-a f}{a b-h^{2}}\))
Equation of the diagonal is y = \(\frac{g h-a f}{h f-b g}\) . x
y(hf – bg) = x(gh – af)
(gh – af) x – (hf- bg) y = 0

A and B are points on both this pair of lines.
Combined equation of AB is
2gx + 2fy + c = 0
OACB is a rhombus
OC and AB are perpendicular
2g(gh – af) – 2f(.hf – bg) = 0
hg² – afg – hf² + bfg = 0
(a – b) fg + h(f² – g²) = 0

Question 20.
If two of the sides of a parallelogram are represented by ax² + 2hxy + by² = 0 and px + qy = 1 is one of its diagonals, prove that the other diagonal is y(bp – hq) = x(aq – hp).
Solution:
Let OACB be the parallelogram two of whose sides \(\overleftrightarrow{O A}\), \(\overleftrightarrow{O B}\) are represented by the equation H ≡ ax² + 2hxy + by² = 0. Since the other pair of sides \(\overleftrightarrow{A C}\) and \(\overleftrightarrow{B C}\) are respectively parallel to \(\overleftrightarrow{O B}\) and \(\overleftrightarrow{O A}\), their combined equation will be of the form S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0. Then the equation of the diagonal \(\overleftrightarrow{A B}\) is 2gx + 2fy + c = 0. But this line is given to be px + qy = 1 (or) -pcx – qcy + c = 0 where c ≠ 0

Therefore 2g = – pc, 2f = – qc …………….. (1)
The vertex C of the parallelogram = (\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g f-a f}{a b-h^{2}}\))
∴ Equation of the diagonal \(\overleftrightarrow{O C}\) is
(gh – af) x = (hf- bg)y
i.e., c(-ph + aq) x = c(-hq + bp)y (By 1)
(or) (aq – hp) x = (bp – hq) y (since c ≠ 0)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B The Straight Line Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B The Straight Line Important Questions

Question 1.
Find the equation of the straight line passing through the point (2, 3) and making non-zero intercepts on the axes of coordinates whose sum is zero. [Mar 12]
Solution:
Equation of the line in the intercept form is
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Given b = -a
Equation of the line is \(\frac{x}{a}\) – \(\frac{y}{a}\) = 1
⇒ x – y = a
This line passes through (2, 3)
2 – 3 = a ⇒ a = -1
Equation of the line is
x – y = -1 or x – y + 1 = 0

Question 2.
Find the equation of the straight line passing through the points (at₁², 2at₁ ) and (at₂², 2at₁ ).
Solution:
Equation of the given points.
(x – x₁) (y – y₂) = (y – y₁) (x₁ – x₂)
( x -at₁²) (2at₁ – 2at₂)
= (y – 2at₁) (at, — at2)
(x – at₁²) .2a(t₁ – t₂) = (y – 2at₁) a. (t₁² – t₂²)
2x – 2at₁² = y(t₁ + t₂) – 2at₁² + 2at₁t₂ = 0
2x – (t₁ + t₂)y + 2at₁t₂ = 0

Question 3.
Find the equation of the straight line passing through A (-1, 3) and i) parallel ii) perpendicular to the straight line passing through B(2, – 5) and C(4, 6). [Mar 11]
Solution:
Slope of BC = \(\frac{-5-6}{2-4}\) = \(\frac{-11}{-2}\) = \(\frac{11}{2}\)
i) The required line is parallel to BC and passes through A(- 1, 3)
Equation of the parallel line is
y – 3 = \(\frac{11}{2}\)(x + 1)
2y – 6 = 11x + 11
11x – 2y + 17 = 0

ii) The required line is perpendicular to BC.
Slope of the required line = –\(\frac{1}{m}\) = –\(\frac{2}{11}\)
This line passes through A (-1, 3)
Equation of the required line is
y – 3 = –\(\frac{2}{11}\)(x + 1)
11y – 33 = -2x – 2
2x + 11y – 31 = 0

Question 4.
Prove that the points (1, 11), (2, 15) and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
A(1, 11), B(2, 15) and C (-3, -5) are the given points.
Equation of AB is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y – 11) (1 -2) = (x – 1) (11 – 15)
-(y – 11) = -4 (x – 1)
-y + 11 = – 4x + 4
4x – y + 7 = 0
C (-3, -5) .
4x – y + 7 = 4(-3) + 5 + 7
= -12 + 12 = 0
C lines on AB ⇒ A, B, C are collinear.
Equation of the line containing them is 4x – y + 7 = 0

Question 5.
A straight line passing through A(1, -2) makes an angle tan^-1 \(\frac{4}{3}\) with the positive direction of the X – axis in the anticlock-wise sense. Find the points on the straight line whose distance from A is 5.
Solution:

Given α = tan^-1 \(\frac{4}{3}\) ⇒ tan α = \(\frac{4}{3}\)

cos α = \(\frac{3}{5}\), sin α = \(\frac{4}{5}\)
(x₁, y₁) = (1, -2) = x₁ = 1, y₁ = -2
Case i) .
r = 5
x = x₁ + r cos α = 1 + 5 . \(\frac{4}{5}\) = 1 + 4 = 5
y = y₁ + r sin α= – 2 + 5 . \(\frac{3}{5}\) = -2 + 3 = 1
Co-ordinates of B are (5, 1)

Case ii) r = -5
x = x₁ + r cos α = 1 – 5 . \(\frac{4}{5}\) = 1 – 4 = -3
y = y₁ + r sin α = -2 – 5 . \(\frac{3}{5}\) = -2 – 3 = -5
Co-ordinates of C are (- 3, – 5)

Question 6.
A straight line parallel to the line y = \(\sqrt{3} x\) passes through Q(2, 3) and cuts the line 2x + 4y – 27 = 0 at P. Find the length of PQ.
Solution:
PQ is parallel to the straight line y = \(\sqrt{3} x\)
tan α = \(\sqrt{3}\) = tan 60°
α = 60°
Q(2, 3) is a given point

Co-ordinates of any point P are
(x₁ + r cos α, y₁ + r sin α)
(2 + r cos 60°, 3 + r sin 60°)

Question 7.
Transform the equation
3x + 4y+ 12 = 0 into
i) slope—intercept form
ii) Intercept form and
in) normal form
Solution:
The given equation is 3x + 4y + 12=0
i) Slope-intercept form
4y = -3x – 12
y = (-\(\frac{3}{4}\))x + (-3)
Slope = –\(\frac{3}{4}\) , y – intercept = -3.

ii) Intercept form
-3x – 4y = 12
–\(\frac{3 x}{12}\) – \(\frac{4 y}{12}\) = 1
\(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
x – intercept = -4, y – intercept = -3

iii) Normal form
-3x – 4y = 12
Dividing with \(\sqrt{9+16}\) = 5
(-\(\frac{3}{5}\))x + (-\(\frac{4}{5}\))y = \(\frac{12}{5}\)
Let cos α = \(\frac{-3}{5}\) and sin α = –\(\frac{4}{5}\)
p = \(\frac{12}{5}\) so that
x cos α + y sin α = p
α lies in third quadrant so that
α = π + tan^-1(\(\frac{4}{3}\))

Question 8.
If the area of the triangle formed by the straight line x = 0, y = 0 and 3x + 4y = a (a > 0) is 6, find the value of a. [May 11]
Solution:
Equation of the line is 3x + 4y = a

Given \(\frac{a^{2}}{2 a}\) = 6 ⇒ a² = 144
a = ±12
But a > 0
a = 12

Question 9.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y -13 = 0 and 8x – 11y – 33 = 0 are concurrent.
Solution:
Let L₁, L₂, L₃ be the straight lines whose equations are respectively
2x – 3y + k = 0 ………………. (1)
3x – 4y – 13 = 0 ……………….. (2)
8x – 11y – 33 = 0 ………………. (3)
Solving (2) and (3) for x and y

Point of the lines (2) and (3) is (11, 5)
The given lines L₁, L₂, L₃ are concurrent
∴ L₁ contain (11, 5)
∴ 2(11) – 3(5) + k = 0
k = -7

Question 10.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a³ + b³ + c³ = 3abc.
Solution:
The equation of the given lines are
ax + by +c = 0 ………………… (1)
bx + cy + a = 0 ……………….. (2)
cx + ay + b = 0 ……………….. (3)
Solving (1) and (2) point of intersection is got by

c(ab – c²) +a (bc – a²) + b(ca – b²) = 0
abc – c³ + abc – a³ + abc – b³ = 0
∴ a³ + b³ + c³ = 3abc.

Question 11.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1 meets the co-ordinate axes at A and B. Show that the locus of the mid point of \(\overline{\mathrm{AB}}\) is 2(a + b) xy = ab(x + y)
Solution:
Equations of the given lines are \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1
Solving the point of intersection P(\(\frac{a b}{a+b}\), \(\frac{a b}{a+b}\))
Q (x₀, y₀) is any point on the locus
⇔ The line with x – intercept 2x₀, y – intercept 2y₀, passes through P
⇔ P lies on the straight line \(\frac{x}{2 x_{0}}\) + \(\frac{y}{2 y_{0}}\) = 1
i.e., \(\frac{a b}{a+b}\)(\(\frac{1}{2 x_{0}}\) + \(\frac{1}{2 y_{0}}\)) = 1
⇒ \(\frac{a b}{a+b}\) . \(\frac{x_{0}+y_{0}}{2 x_{0} y_{0}}\) = 0
ab(x₀ + y₀) = 2(a + b) x₀y₀
Q(x₀, y₀) lies on the curve
2(a + b)xy = ab(x + y)
Locks the mid of point of
AB is 2(a + b)xy = ab(x +y)

Question 12.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
a, b, c are in A.P.
2b = a + c
a – 2b + c = 0
a.1 + b(-2) + c = 0
Each number of family of straight lines
ax + by + c = 0
passes through the fixed point (1,-2)
∴ Set of lines ax + by + c = 0 for parametric values of a, b and c is a family of concurrent lines.
∴ Point of concurrency is (1, -2).

Question 13.
Find the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°.
Solution:

Squaring and cross multiplying
2(4k + 5)² = 17(k² + 25)
2(16k² + 40k + 25) = 17k² + 425
32k² + 80k + 50 = 17k² + 425
15k² + 80k – 375 = 0
3k² + 16k – 75 = 0
(k – 3) (3k + 25) = 0
k = 3 or -25/3

Question 14.
Find the equation of the straight line passing through (x₀, y₀) and (i) parallel (ii) perpendicular to the straight line
ax + by + c = 0.
Solution:
Equation of the given line is ax + by + c = 0
i) Equation of the parallel line is ax + by = k ………………… (1)
This line passes through P(x₀, y₀) ⇒ ax₀ + by₀ = k ……………… (2)
Subtracting (2) from (1) equation of the required line is a(x – x₀) + b(y – y₀) = 0

ii) Equation of the perpendicular line is bx – ay = k.
This line passes through P(x₀, y₀) ⇒ bx₀ – ay₀ = k
Subtracting, equation of the required line is b(x – x₀) – a(y – y₀) = 0

Question 15.
Find the equation of the straight line perpendicular to the line 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6.
Solution:
Given lines are L₁ = 2x + 3y – 1 = 0
L₂ = 3x + 4y – 6 = 0
Equation of the line passing through the intersection of L₁ = 0, L₂ = 0 is
L₁ + kL₂ = 0
(2x + 3y – 1) + k(3x + 4y – 6) = 0
(2 + 3k)x + (3 + 4k)y – (1 + 6k) = 0 …………………. (1)
This line is perpendicular to 5x – 2y = 7 ………………… (2)
a₁a₂ + b₁b₂ = 0
5(2 + 3k) – 2(3 + 4k) = 0
10 + 15k – 6 – 8k = 0
7k = -4 ⇒ k = – 4/7
Substituting in (1) equation of the required lines
(2 – \(\frac{12}{7}\))x + (3 – \(\frac{16}{7}\))x – (1 – \(\frac{24}{7}\)) = 0
\(\frac{2}{7}\)x + \(\frac{5}{7}\)y + \(\frac{17}{7}\) = 0 ⇒ 2x + 5y + 17 = 0

Question 16.
If 2x – 3y – 5 = 0 is the perpendicular bi-sector of the line segment joining (3, – 4) and (α, β) find α + β.
Solution:
(α, β) is the reflection of (3, -4) is the line
2x – 3y – 5 = 0
\(\frac{\alpha-3}{2}\) = \(\frac{\beta+4}{-3}\) = \(\frac{-2(6+12-5)}{4+9}\) = -2
α – 3 = -4 ⇒ α = -1
β + 4 = 6 ⇒ β = 2
α + β = -1 + 2 = 1

Question 17.
If the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram, show that the area of the parallelogram so formed is \(\left|\frac{(p-a)(r-s)}{b c-a d}\right|\).
Solution:
Let L₁, L₂, L₃, L₄ be the lines given by
L₁ = ax + by + p = 0
L₂ = ax + by + q = 0
L₃ = cx + dy + r = 0
L₄ = cx + dy + s = 0
L₁ and L₂ are parallel: L₃ and L₄ are parallel
Area of the parallelogram = \(\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\sin \theta}\)
d₁ = distance between L₁ and L₂ = \(\frac{|p-a|}{\sqrt{a^{2}+b^{2}}}\)
d₂ = distance between L₃ and L₄ = \(\frac{|r-s|}{\sqrt{c^{2}+d^{2}}}\)

Question 18.
The hypotenuse of a right angled isosceles triangle has its ends at the points (1, 3) and (-4, 1). Find the equations of the legs of the triangle.
Sol:
LetA =(1, 3)and B = (-4, 1) and ABC b a right isosceles triangle with \(\stackrel{\leftrightarrow}{A B}\) as hypotenuse.
We require, therefore, the equations of
\(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\)
Slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{1-3}{-4-1}\) = \(\frac{2}{5}\)
Since the slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\), neither \(\frac{2}{5}\) nor \(\stackrel{\leftrightarrow}{A C}\) is vertical.

Taking the slope of \(\stackrel{\leftrightarrow}{A C}\) as \(\frac{7}{3}\), the slope of \(\stackrel{\leftrightarrow}{B C}\) would be –\(\frac{3}{7}\). Therefore, the equations of \(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\) are respectively.
y – 3 = \(\frac{7}{3}\) (x – 1) and y – 1 = – \(\frac{3}{7}\) (x + 4),
which become 7x – 3y + 2 = 0 and 3x + 7y + 5 = 0
If the lines drawn through A and B respectively parallel to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{A C}\) meet at D, then ∆ABD is also right isosceles, having \(\stackrel{\leftrightarrow}{A B}\) as its hypotenuse.
Therefore, the equations of \(\stackrel{\leftrightarrow}{A D}\) and \(\stackrel{\leftrightarrow}{B D}\) are respectively,
3(x – 1) + 7(y – 3) = 0 and 7(x + 4) – 3(y – 1) = 0
⇒ 3x + 7y – 24 = 0 and 7x – 3y + 31 = 0.
Therefore, the two pairs of legs required are
7x – 3y + 2 = 0, 3x + 7y + 5 = 0 and
3x + 7y – 24 = 0, 7x – 3y + 31 = 0.
Note : ADBC is square.

Question 19.
A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is . bisected at the point (1, 5). Obtain its equation.
Solution:
Let the required line meet 3x + 4y – 4 = 0 at A and 5x – y + 4 = 0 at B, so that AB is the segment between the given lines, with its mid-point at C = (1, 5).
The equation 5x – y + 4 = 0 can be written as
y = 5x + 4 so that any point on \(\stackrel{\leftrightarrow}{B X}\) is (t, 5t + 4) for all real t.
∴ B = (t, 5t + 4) for some t. Since (1, 5) is the mid-point of \(\stackrel{\leftrightarrow}{A B}\)
A = [2 – t, 10 – (5t + 4)]
= [2 – t, 6 – 5t]
Since A lies on 3x + 4y – 4 = 0,
3(2 – 1) + 4(6 – 5t) – 4 = 0
⇒ -23t + 26 = 0
⇒ t = \(\frac{26}{23}\)

equation of \(\stackrel{\leftrightarrow}{A B}\) is y – 5 = \(\frac{107}{3}\) (x – 1)
⇒ 3y – 15 = 107x – 107
⇒ 107x – 3y – 92 = 0.

Question 20.
An equilateral triangle has its incentre at the origin and one side as x + y – 2 = 0. Find the vertex opposite to x + y – 2 = 0.
Solution:
Let ABC be the equilateral triangle and
x + y – 2 = 0 represent side \(\stackrel{\leftrightarrow}{B C}\).
Since O is the incentre of the triangle, \(\stackrel{\leftrightarrow}{A D}\) is the bisector of ∠BAC .
Since the triangle is equilateral, \(\stackrel{\leftrightarrow}{A D}\) is the perpendicular bisector of \(\stackrel{\leftrightarrow}{B C}\).
Since O is also the centroid, AO : OD = 2 : 1. [The centroid, circumcentre incentre and orthocentre coincide]
Let D = (h, k).
Since D is the foot of the perpendicular from O onto \(\stackrel{\leftrightarrow}{B C}\), D is given by

∴ x₁ = -2, y₁ = -2.
∴ A = (-2, -2), the required vertex.

Question 21.
Find the orthocentre of the triangle whose vertices are (-5,-7) (13, 2) and (-5, 6)
Solution:
Let A(-5, -7), B(13,2) and C (-5, 6) be the vertices of a triangle. Let \(\stackrel{\leftrightarrow}{A D}\) be the perpendicular drawn from A to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{B E}\) be the perpendicular drawn from B to \(\stackrel{\leftrightarrow}{A C}\).

Now slope of \(\stackrel{\leftrightarrow}{B C}\) = \(\frac{6-2}{-5-13}\) = \(\frac{-2}{9}\)
Since \(\stackrel{\leftrightarrow}{A D}\) ⊥ \(\stackrel{\leftrightarrow}{B C}\) , slope of \(\stackrel{\leftrightarrow}{A D}\) = \(\frac{9}{2}\) and so, the equation of \(\stackrel{\leftrightarrow}{A D}\) is
9x – 2y = – 45 + 14 = -31 ………………… (1)
Equation of \(\stackrel{\leftrightarrow}{A C}\) is x = -5 which is a vertical line and therefore equation of \(\stackrel{\leftrightarrow}{B E}\) is y = 2. ……………… (2)
Point of intersection of the lines (1) and (2) is (-3, 2) which is the orthocentre of ∆ ABC.

Question 22.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0 and x + y + 2 = 0, find the orthocentre of the triangle.
Solution:
Let ∆ ABC be the given triangle
Let the equations x – 2y + 5 = 0 ……………… (1)
7x + y – 10 = 0 ……………… (2)
and x + y + 2 = 0 ……………….. (3)
represent the sides \(\overleftrightarrow{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{BC}}\), and \(\overleftrightarrow{\mathrm{CA}}\)

Let \(\overleftrightarrow{\mathrm{AD}}\) and \(\overleftrightarrow{\mathrm{BE}}\) be the altitudes drawn from A and B respectively to the sides \(\overleftrightarrow{\mathrm{BC}}\) & \(\overleftrightarrow{\mathrm{CA}}\)
Solving the equations (1) and (3), we obtain A =(-3,1).
Since \(\overleftrightarrow{\mathrm{AD}}\) ⊥ \(\overleftrightarrow{\mathrm{BC}}\), the equations of \(\overleftrightarrow{\mathrm{AD}}\) is
x – 7y = -3 – 7 = -10 …………………… (4)
Solving the equation (1) and (2), we obtain B = (1, 3).
Since \(\overleftrightarrow{\mathrm{BE}}\) ⊥ \(\overleftrightarrow{\mathrm{AC}}\) , the equation of BE is
x – y = 1 – 3 = -2 …………… (5)
Point of intersection of the lines (4) and (5) is H(\(\frac{-2}{3}\), \(\frac{4}{3}\)) which is the orthocentre of ∆ ABC.

Question 23.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5) and (5, -1).
Solution:
Let the vertices of the triangle be
A(1, 3), B(-3, 5) and (5,-1).
The midpoints of the sides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\) are respectively D(1, 2) and E(3, 1).
Let S be the point of intersection of the perpendicular bisectors of the sides \overline{\mathrm{BC}} and \overline{\mathrm{CA}}

Since the slope of \(\overleftrightarrow{\mathrm{BC}}\) = \(\frac{5+1}{-3-5}\) = \(\frac{-3}{4}\), the slope \(\overleftrightarrow{\mathrm{SD}}\) is \(\frac{4}{3}\) and therefore its equation is 4x – 3y = 4 – 6 = -2 …………….. (1)
Slope of \(\overleftrightarrow{\mathrm{AC}}\) = \(\frac{3+1}{1-5}\) = -1 ⇒ Slope of –\(\sqrt{3x}\) = 1
∴ Equation of \(\overleftrightarrow{\mathrm{SE}}\) is x – y = 3 – 1 = 2 ………………….. (2),
Solving the equations (1) and (2),we obtain S = (-8, -10) which is the circumcentre of ∆ ABC.

Question 24.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May 11, 05; Mar. 06]
Solution:
Let the given equations 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0 represent the sides \(\overleftrightarrow{\mathrm{BC}}\), \(\overleftrightarrow{\mathrm{CA}}\) and \(\overleftrightarrow{\mathrm{AB}}\) respectively of ∆ ABC.
Solving the above equations two by two, we obtain the vertices A(-2, 3), B(1, -2) and (2, 1) of the given triangle.
The midpoints of the sides \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{CA}}\) are respectively D = (\(\frac{3}{2}\), \(\frac{-1}{2}\)) and E = (0, 2).

Equation of \(\overleftrightarrow{\mathrm{SD}}\) the perpendicular bisector of \(\overline{\mathrm{BC}}\) is x + 3y = 0 and the equation of \(\overleftrightarrow{\mathrm{SE}}\) the perpendicular bisector of \(\overline{\mathrm{AC}}\) is 2x – y + 2 = 0. Solving these two equations, we obtain the point of intersection of the lines \(\overleftrightarrow{\mathrm{SD}}\) and \(\overleftrightarrow{\mathrm{SE}}\) which is S(\(\frac{-6}{7}\), \(\frac{2}{7}\)) the circumcentre of ∆ ABC.

Question 25.
Find the incentre of the triangle formed by the straight lines y = \(\sqrt{3}\)x, y = –\(\sqrt{3}\)x and y = 3.
Solution:

The straight lines y = \(\sqrt{3x}\) and y = –\(\sqrt{3x}\) respectively make angles 60° and 120° with the positive direction of X – axis.
Since y = 3 is a horizontal line, the triangle formed by the three given lines is equilateral. So in-centre is same and centroid.
Vertices of the triangle and O(0, 0), A(\(\sqrt{3}\), 3) and D(\(\sqrt{3}\), 3).
∴ Incentre is (\(\frac{0+\sqrt{3}-\sqrt{3}}{3}\), \(\frac{0+3+3}{3}\)) = (0, 2)

Question 26.
Find the equation of the straight line whose distance from the origin is 4, If the normal ray from the origin to the straight line makes an angle of 135° with the positive direction of the x-axis.
Solution:
The equation of the given line is
x cos α + y sin α = p where p = 4 and α = 135°
∴ x(\(\frac{-1}{\sqrt{2}}\)) + y(\(\frac{1}{\sqrt{2}}\)) = 4 ⇒ or x – y + 4\(\sqrt{2}\) = 0

(i). Transform the equation x + y + 1 = 0 in to normal from.
Solution:
x + y + 1 = 0
⇔ (\(\frac{-1}{\sqrt{2}}\))n + (\(\frac{-1}{\sqrt{2}}\)) y = \(\frac{1}{\sqrt{2}}\)
⇔ x cos \(\frac{5\pi}{42}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)
Hence the normal form of the equation of the given straight line is x cos \(\frac{5\pi}{4}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\) and the distance of the line from the origin is = \(\frac{1}{\sqrt{2}}\)

ii) A straight line passing through A(1, -2) makes and angle Tan^-1\(\frac{4}{3}\) with the positive direction of the x – axis in the anti-clock wise sense. Find the points of the straight line whose distance from A is 5.
Solution:
The paramatic equation of the lines A (1, -2)
Slope is \(\frac{4}{3}\) (∵ tan θ = \(\frac{4}{3}\))
are x = 1 + r cos θ = 1 + r(\(\frac{3}{5}\)) and y
= -2 + r sinθ
= -2 + r (\(\frac{4}{5}\))
The points on the above line a + a distance of (r) = 5 correspond to r = ±5 in the above equation and are
∴ (4, 2) and (-2, -6)

Question 27.
Trans form the equation 3x + 4y + 12 = 0 in to (i) slope – intercept from (ii) intercept form (iii) normal form.
Solution:
(i) slop – intercept form
3x + 4y + 12 = 0
⇔ 4y = -3x – 12
⇔ y = (\(\frac{-3}{4}\)) x + (-3)
∴ Slope = (\(\frac{-3}{4}\)) and y – intercept = -3,

ii) Intercept form :
3x = 4y + 12 = 0
⇔ \(\frac{-3y}{12}\) – \(\frac{-4y}{12}\) = 1
⇔ \(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
∴ x-intercept of the line is -4, and tan y-intercept is -3.

iii) Normal form
3x = 4y + 12 = 0
⇔ -3x + -4y – 12 = 0
⇔ (\(\frac{-3}{5}\))x + (\(\frac{-4}{5}\))y = \(\frac{12}{5}\)
⇔ x cos α + y sin α = p ⇒ p\(\frac{12}{5}\) and cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\) determine the angle α in (0, 2π).

Question 28.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
The angle between the given lines
= cos^-1 \(\frac{-6+1}{\sqrt{5 \times 10}}\)
= cos^-1 [latex]\frac{5}{\sqrt{2}}[/latex] = cos^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

Question 29.
Find Q(h,k) in the foot of the perpendicular from p(x₁, y₁) on the straight lines ax + by + c = 0 then (h – x₁) ; a = (k – y₁); b = -(ax₁ + by₁ + c); (a² + b²)
Solution:
Equation of \(\stackrel{\leftrightarrow}{P Q}\) which is normal to the given straight line
L : ax + by + c = 0
bx – ay = bx₁ – ay₁

∴ Q ∈ \(\stackrel{\leftrightarrow}{P Q}\) we have
bh – ak = bx₁ – ay₁
∴ b(h – x₁) = a(k – y₁)
or (h – x₁) a = (k – y₁); b.
But, this implus that h = aλ + x₁, and k = bλ + y₁
for some λ ∈ R. sine Q(h, k) in point on L.
a(aλ + x₁) + b(bλ + y₁) + c = 0
i.e., λ = \(\frac{\left(a x_{1}+b y_{1}+c\right.}{\left(a^{2}+b^{2}\right)}\)
∴ (h – x₁) ; a = (k – y₁); b
= – (ax₁ + by₁ + c); (a² + b²)

Question 30.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y -1 = 0.
Solution:
The equation of the given straight lines is 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0.
formula : \(\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\) = \(\frac{-6+1}{\sqrt{6^{2}+8^{2}}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 31.
Find the condition for the points (a, o) (h, k) and (o, b) when ab ≠ 0 to be collinear. [Mar 10]
Solution:
The equation of the line passing through (a, o) and (o, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
The given points are collinear ⇒ (h, k) lies on the above line ⇒ \(\frac{h}{a}\) + \(\frac{k}{b}\) = 1
⇒ hb + ka = ab

Question 32.
Find the area of the triangle formed by the straight lines x cos a + y sin a = p and the axes of co-ordinates. [Mar 10]
Solution:
The area of the triangle formed by the line ax + by + c = 0
and the co-ordinate axes is \(\frac{c^{2}}{2|a b|}\)
∴ Area of the triangle = \(\frac{p^{2}}{2|\cos \alpha \cdot \sin \alpha|}\)
= \(\frac{p^{2}}{|\sin 2 \alpha|}\)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Transformation of Axes Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Transformation of Axes Important Questions

Question 1.
When the origin is shifted to (2, 3) by the translation of axes, the coordinates of a point p are changed as (4, -3). Find the coordinates of P in the original system.
Solution:
(h, k) = (2, 3) ⇒ h = 2, k = 3
(x’, y’) = (4, 3) ⇒ x = 4, y = -3
x = x’ + h = 4 + 2 = 6, y = y’ + k = -3 + 3 = 0
original Co-ordinates are (6, 0)

Question 2.
Find the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0, where h² ≠ ab.
Solution:
Let the origin be shifted to (α, β) by the translation of axes.
Then x = x’ + α,
y = y’ + β
On substituting these in the given equation, we get
a(x’ + α)² + 2h(x’ + α) (y’ + β) + b(y’ + β)² + 2g(x’ + α) + 2f(y’ + β) + c = 0
Which gives
ax’² + 2hx’y’ + by² + 2x'(α + β + g) + 2y'(hα + bβ + f) + aα² + 2hαβ + bβ² + 2gα + 2fβ + c = 0 ……….. (1)
If equation (1) has to be free from the first degree terms, then we have
aα + hβ + g = 0 and hα + bβ +1 = 0
Solving these equations for a and 13, we get
α =\(\frac{h f-b g}{a b-h^{2}}\) , β = \(\frac{g h-a f}{a b-h^{2}}\)
Therefore, the origin is to be shifted to
\(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Question 3.
Find the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
ax² + by² + 2gx + 2fy + c = 0, where a ≠ 0, b ≠ 0.
Solution:
Here the given equation does not contain xy term. Hence writing h = 0 in the result of
Problem 12, the required point is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

Question 4.
If the point P changes to (4, -3) when the axes are rotated through an angle of 135°, find the coordinates of P with respect to the original system.
Solution:
Here (x’, y’) = (4, -3); θ = 135°
Let (x, y) be the coordinates of P, then
x = x’ cos θ – y’ sin θ
= 4cos 135° – (-3) sin 135°

Therefore, the coordinates of P with respect to the original system are (\(\frac{-1}{\sqrt{2}}\), \(\frac{7}{\sqrt{2}}\))

Question 5.
Show that the axes are to be rotated through an angle of \(\frac{1}{2}\) Tan^-1(\(\frac{2 h}{a-b}\)) so as to remove the xy term from the equation ax² + 2hxy + by² = 0, if a ≠ b and through the angle \(\frac{\pi}{4}\), if a = b. [Mar 13, 06]
Solution:
If the axes are rotated through an angle ‘θ’,
then
x = x’ cos θ – y’ sin θ
y = x’ sin θ + y’ cos θ
Therefore the given equation transforms as
a(x’ cos θ – y’ sin θ)² + 2h(x’ cosθ – y’ sinθ) (x’ sinθ + y’ cosθ) + b(x’ sin θ + y’ cos θ)² = o
To remove x’y’ term from the equation, the coefficient of x’y’ term must be zero.
So, (b – a) sin θ cos θ + h(cos²θ – sin²θ) = 0
i.e., h cos 2θ = \(\frac{a-b}{2}\) sin 2θ
i.e., tan 2θ = \(\frac{2 h}{a-b}\), if a ≠ b and h cos 2θ = 0, if a = b
Therefore θ = \(\frac{1}{2}\) Tan^-1(\(\frac{2 h}{a-b}\)), if a ≠ b and θ = \(\frac{\pi}{4}\), If a = b

Question 6.
When the origin is shifted to (-2, -3) and the axes are rotated through an angle 45° find the transformed equation of 2x² + 4xy – 5y² + 20x – 14 = 0.
Solution:
Here (h, k) = (-2, -3), h = -2, k = -3
θ = 45°
Let (x’, y’) be the new co-ordinates of any point (x, y) is the plane after transformation
x = x’ cos θ – y’ sin θ + h = -2 + x’ cos 45° – y’ sin 45°.
= -2 + \(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\)
y = x’ si nθ + y’ cos θ + k = x’ sin 45° + y’ cos
45° – 3 = -3 + \(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\)
The transformed equation is

⇒ (x’ – y’)² + 8 – 4\(\sqrt{2}\) (x’ – y’) + 2(x’² – y’²) – 6\(\sqrt{2}\) (x’ – y’) – 4\(\sqrt{2}\) (x’ + y’) + 24 = 0
⇒ –\(\frac{5}{2}\) (x’ – y’)² – 45 + 15\(\sqrt{2}\) (x’ + y’) + 10\(\sqrt{2}\) (x’ – y’) – 40 – 11\(\sqrt{2}\) (x’ + y’) + 66 – 14 = 0
⇒ x’² + y’² – 2x’y’ + 2x’² – 2y’² – \(\frac{5}{2}\) (x’² + y’² + 2x’y’) – 1 = 0
\(\frac{1}{2}\) x’² – \(\frac{7}{2}\) y’² – 7x’y’ – 1 = 0
i.e., x’² – 7y’² – 14x’y’ – 2 = 0
The transformed equation is (dropping dashes)
x² – 7y² – 14xy – 2 = 0

Question 7.
When the origin is shifted to (-2, 3) by translation of axes, let us find the co-ordinates of (1, 2) with respect to new axes.
Solution:
Here (h, k) = (-2, 3)
Let (x, y) = (1, 2) be shifted to (x’ y’) by the translation of axes.
Then (x’, y’) = (x – h, y – k)
= (1 – (-2), 2 – 3) = (3, -1)

Question 8.
When the origin is shifted to (3, 4) by the translation of axes, let us find the transformed equation of 2x² + 4xy + 5y² = 0.
Solution:
Here (h, k) = (3, 4)
on substituting x = x’ + 3 and y = y’ + 4 in the given equation.
as per note the equation f(x, y) = 0 of the curve is transformed as f(x’ + h, y’ + k) = 0]
we get
2(x’ + 3) + 4(x’ + 3) (y’ + 4) + 5(y’ + 4)² = o
Simplifying this equation, we get
2x’² + 4x’y’ + 5y’² + 28x’ + 52y’ + 146 = 0
This equation can be written (dropping dashes) as
2x² + 4xy + 5y² + 28x + 52y + 146 = 0

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Locus Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Locus Important Questions

Question 1.
Find the equation of the locus of a point which is at a distance 5 from (-2, 3) in the xoy plane.
Solution:
Let the given point be A = (-2, 3) and P(x, y) be a point on the plane.
The geometric condition to be satisfied by P to be on the locus is that
AP = 5 …………… (1)
Expressing this condition algebraically, we get
\(\sqrt{(x+2)^{2}+(y-3)^{2}}\) = 5
i.e., x² + 4x + 4 + y² – 6y + 9 = 25
i.e., x² + y² + 4x – 6y – 12 = 0 …………….. (2)
Let Q(x₁, y₁) satisfy (2).
Then, x₁² + y₁² + 4x₁ – 6y₁ – 12 = 0 …………. (3)
Now the distance of A from Q is
AQ = \(\sqrt{\left(x_{1}+2\right)^{2}+\left(y_{1}-3\right)^{2}}\)
∴ AQ² = x₁² + 4x₁ + 4 + y₁² – 6y₁ + 9
= (x₁² + y₁² + 4x₁ – 6y₁ – 12) + 25
= 25 (by using (3))
Hence AQ = 5.
This means that Q(x₁, y₁) satisfies the geometric condition (1).
∴ The required equation of locus is
x² + y² + 4x – 6y – 12 = 0.

Question 2.
Find the equation of locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
Solution:
Let P(x, y) he a point on the locus. Then the geometric condition to be satisfied by P is
PA = 2PB …………….. (1)
i.e., PA² = 4PB²
i.e., (x – 3)² + y² = 4[(x + 3)² + y²]
i.e., x² – 6x + 9 + y² = 4(x² + 6x + 9 + y²]
i.e., 3x² + 3y² + 30x + 27 = 0
i.e., x² + y² + 10x + 9 = 0 ………………. (2)
i.e., Q(x₁, y₁) satisfy (2).
Then x₁² + y₁² + 10x₁ + 9 = 0 …………….. (3)
Now QA² = (x₁ – 3)² + y₁²
= x₁² – 6x₁ + 9 + y₁²
= 4x₁² + 24x₁ + 36 + 4y₁² – 3x₁² – 30x₁ – 27 – 3y₁²
= 4(x₁² + 6x₁ + 9 + y₁²) – 3(x₁² + 10x₁ + 9 + y₁²)
= 4(x₁² + 6x₁ + 9 + y₁²) (by using(3))
= 4 [(x₁ + 3)² + y₁²]
= 4QB²
∴ QA = 2QB.
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
x² + y² + 10x + 9 = 0.

Question 3.
Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).
Solution:
Let A = (4, 0) and B = (0, 4).
Let P(x, y) be a point such that. PA and PB are perpendicular. Then PA² + PB² = AB².
i.e., (x – 4)² + y² + x² + (y – 4)² = 16 + 16
i.e., 2x² + 2y² – 8x – 8y = 0
or x² + y² – 4x – 4y = 0
Let Q(x₁, y₁) satisfy (2) and Q be different from A and B.
Then x₁² + y₁² — 4x₁ – 4y₁ = 0,
(x₁, y₁) ≠ (4, 0) and (x₁, y₁) ≠ (0, 4) ……………… (3)
Now QA² + QB² = (x₁ – 4)² + y₁² + x₁² + (y₁ – 4)²
= x₁² – 8x₁ + 16 + y₁² + x₁² + y₁² – 8y₁ + 16
= 2(x₁² + y₁² – 4x₁ – 4y₁) + 32
= 32 (by using (3))
= AB²
Hence QA² + QB² = AB², Q ≠ A and Q ≠ B.
This means that Q(x₁, y₁) satisfies (1).
∴ The required equation of locus is (2),which is the circle with \(\overline{\mathrm{AB}}\) as diameter, deleting the points A and B.
Though A and B satisfy equation (2), they do not satisfy the required geometric condition.

Question 4.
Find the equation of the locus of P, if the ratio of the distances from P to A(5, -4) and B(7, 6) is 2 : 3. [Mar 14]
Solution:
Let P(x, y) be any point on the locus.
The geometric condition to be satisfied by P
is \(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{1}{2}\)
i.e., 3AP = 2PB …………….. (1)
i.e., 9AP² = 4PB²
i.e., 9[(x – 5)² + (y + 4)²] = 4[(x – 7)² + (y – 6)²]
i.e., 9[x² + 25 – 10x + y² + 16 + 8y]
= 4[x² + 49 – 14x + y² + 36 – 12y]
i.e., 5x² + 59 – 34x + 120y + 29 = 0 ……………….. (2)
Let Q(x₁, y₁) satisfy (2). Then
5x₁² + 5y₁² – 34x₁ + 120y₁ + 29 = 0 ……………….. (3)
Now 9AQ² = 9[x₁² + 25 – 10x₁ + y₁² + 16 + 8y₁]
= 5x₁² + 5y₁² – 34x₁ + 120y₁ + 29 + 4x₁² + 4y₁² – 56x₁ – 48y₁ + 340
= 4[x₁² + y₁² – 14x₁ – 12y₁ + 49 + 36] (by using (3))
= 4[(x₁ – 7)² + (y₁ – 6)²] = 4PB²
Thus 3AQ = 2PB. This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
5(x² + y²) – 34x + 120y + 29 = 0.

Question 5.
A(2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5. [Mar 11]
Solution:
Let P(x, y) be a point on the locus.
The geometric condition to be satisfied by P is that,
area of ∆PAB = 8.5 …………………(1)
i.e., \(\frac{1}{2}\)|x(3 – 4) + 2(4 – y) – 3(y – 3)| = 8.5
i.e., |-x + 8 – 2y – 3y + 9| = 17
i.e., |-x – 5y + 17| = 17
i.e., -x – 5y + 17 = 17 or -x – 5y + 17 = -17
i.e., x + 5y = 0 or x + 5y = 34
∴ (x + 5y) (x + 5y – 34) = 0
i.e., x² + 10xy + 25y² – 34x – 170y = 0 …………… (2)
Let Q(x₁, y₁) satisfy (2). Then
x₁ + 5y₁ = 0 or x₁ + 5y₁ = 34 ……………………. (3)
Now, area of ∆QAB
= \(\frac{1}{2}\)|x₁(3 – 4) + 2(4 – y₁) – 3(y₁ – 3)|
= \(\frac{1}{2}\)|-x₁ + 8 – 2y₁ – 3y₁ + 9|
= \(\frac{1}{2}\) |-x₁ – 5y₁ + 17|
= \(\frac{17}{2}\) = 8.5 (by using (3))
This means that Q(x₁, y₁) satisfies (1).
Hence, the required equation of locus is
(x + 5y)(x + 5y – 34) = 0 or
x² + 10xy + 25y² – 34x – 170y = 0.