Mahesh

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 4 Pair of Straight Lines to solve questions creatively.

Intermediate 1st Year Maths 1B Pair of Straight Lines Formulas

→ If ax² + 2hxy + by² = 0 represents a pair of lines, then the sum of the slopes of lines is \(-\frac{2 h}{b}\) and the product of the slopes of lines is \(\frac{a}{b}\).

→ If ‘θ’ is an angle between the lines represented by ax² + 2hxy + by² = 0
then cos θ = \(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{a+b}\)
If ‘θ’ is accute, cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\); tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{|a+b|}\)

→ If h² = ab, then ax² + 2hxy + by² = 0 represents coincident or parallel lines.

→ ax² + 2hxy + by² = 0 represents a pair of ⊥^lr lines ⇒ a+ b = 0 i.e., coeft. of x² + coeff. of y² = 0.

→ The equation of pair of lines passing through origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and perpendicular to ax² + 2hxy + by² = 0 isb (x – x₁)² – 2h(x – x₁) (y – y₁) + a(y – y₁)² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is a(x-x,)² + 2h(x-x1)(y-y1) + b(y – y₁)² = 0.

→ The equations of bisectors of angles between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, is \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{\left(a_{2} x+b_{2} y+c_{2}\right)}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

→ The equation to the pair of bisectors of angles between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy.

→ The product of the perpendicular is from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

→ The area of the triangle formed by ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h / m+b\right|^{2} \mid}\)

→ The line ax + by + c = 0 and pair of lines (ax + by)² – 3(bx – ay)² =0 form an equilateral triangle and the area is \(\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)}\) sq.units

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines, then

• abc + 2fgh – af² – bg² – ch² = 0
• h² ≥ ab
• g² ≥ ac
• f² ≥ be

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines and h² > ab, then the point of intersection of the lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel lines then h² = ab and af² = bg²
The distance between the parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}=2 \sqrt{\frac{f^{2}-b c}{b(a+b)}}\)

Pair of Straight Lines:
Let L₁ = 0, L₂ = 0 be the equations of two straight lines. If P(x₁, y₁) is a point on L₁ then it satisfies the equation L₁ = 0. Similarly, if P(x₁, y₁) is a point on L₂ = 0 then it satisfies the equation.

If P(x₁, y₁) lies on L₁ or L₂, then P(x₁,y₁) satisfies the equation L₁L₂= 0.
L₁L₂ = 0 represents the pair of straight lines L₁ = 0 and L₂ = 0 and the joint equation of L₁ = 0 and L₂ = 0 is given by L₁ L₂= 0. ……………(1)
On expanding equation (1) we get and equation of the form ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 which is a second degree (non – homogeneous) equation in x and y.
Definition: If a, b, h are not all zero,then ax² + 2hxy + by² = 0 is the general form of a second degree homogeneous equation in x and y.
Definition: If a, b, h are not all zer, then ax² + 2hxy + by² + 2gx + 2 fy + c = 0 is the general form of a second degree non – homogeneous equation in x and y.

Theorem:
If a, b, h are not all zero and h² ≥ ab then ax² + 2hxy + by² = 0 represents a pair of straight lines passing through the origin.
Proof:
Case (i) : Suppose a = 0.
Given equation ax² + 2hxy + by² = 0 reduces to 2hxy + by² = 0 ^ y(2hx + by) = 0 .
Given equation represents two straight lines y = 0 ………..(1) and 2hx + by = 0 ………(2) which pass through the origin.
Case (ii): Suppose a ≠ 0.
Given equation ax² + 2hxy + by² = 0
⇒ a²x² + 2ahxy + aby² = 0
⇒ (ax)² + 2(ax)(hy) + (hy)² – (h² – ab)y² = 0
⇒ (ax + hy)² – (y\(\sqrt{h^{2}-a b}\))² = 0
[ax + y (h + \(\sqrt{h^{2}-a b}\))][ax + y (h – \(\sqrt{h^{2}-a b}\)] = 0

∴ Given equation represents the two lines
ax + hy + y\(\sqrt{h^{2}-a b}\) = 0, ax + hy – y\(\sqrt{h^{2}-a b}\) = 0 which pass through the origin.

Note 1:

• If h² > ab , the two lines are distinct.
• If h² = ab , the two lines are coincident.
• If h² < ab , the two lines are not real but intersect at a real point (the origin).
• If the two lines represented by ax² + 2hxy + by² = 0 are taken as l₁x + m₁y = 0 and l₂x + m₂y = 0 then
  ax² + 2kxy + by² = (4 x + m₁y) (x + m₂y) = 2 x² + (l₁m₂ + l₂m₁) xy + m₁m₂ y²
• Equating the coefficients of x², xy and y² on both sides, we get l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Theorem:
If ax² + 2hxy + by² = 0 represent a pair of straight lines, then the sum of slopes of lines is \(\frac{-2 h}{b}\) product of the slopes is \(\frac{a}{b}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………….(1) and l₂x + m₂y = 0 ………….(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
Slopes of the lines (1) and (2) are –\(\frac{l_{1}}{m_{1}}\) and \(-\frac{l_{2}}{m_{2}}\).
sum of the slopes = \(\frac{-l_{1}}{m_{1}}+\frac{-l_{2}}{m_{2}}=-\frac{l_{1} m_{2}+l_{2} m_{1}}{m_{1} m_{2}}=-\frac{2 h}{b}\)
Product of the slopes = \(\left(\frac{-l_{1}}{m_{1}}\right)\left(-\frac{l_{2}}{m_{2}}\right)=\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{a}{b}\)

Angle Between A Pair of Lines:
Theorem :
If θ is the angle between the lines represented by ax² + 2hxy + by² = 0, then cos θ = ±\(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁ x + m₁ y = 0 ………..(1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Let θ be the angle between the lines (1) and (2). Then cos θ = ±\(\frac{l_{1} l_{2}+m_{1} m_{2}}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)

Note 1:
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then tan θ = ±\(\frac{2 \sqrt{h^{2}-a b}}{a+b}\) and sin θ = \(\frac{2 \sqrt{h^{2}-a b}}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Conditions For Perpendicular And Coincident Lines:

• If the lines ax² + 2hxy + by² = 0 are perpendicular to each other then θ = π/ 2 and cos θ = 0 ⇒ a + b = 0 i.e., co-efficient of x² + coefficient of y² = 0.
• If the two lines are parallel to each other then 0 = 0.
  ⇒ The two lines are coincident ⇒ h² = ab

Bisectors of Angles:
Theorem:
The equations of bisectors of angles between the lines a₁ x + b₁ y + c₁ = 0, a₂ x + b₂ y + c₂ = 0 are \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\) = ±\(\frac{a_{2} x+b_{2} y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

Pair of Bisectors of Angles:
The equation to the pair bisectors of the angle between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy (or) \(\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……….(1)
and l₂x + m₂y = 0 ………(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The equations of bisectors of angles between (1) and (2) are \(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}-\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0 and
\(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}+\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0

The combined equation of the bisectors is

Theorem
The equation to the pair of lines passing through (x₀, y₀) and parallel ax² + 2hxy + by² = 0
is a( x – x₀)² + 2h( x – x₀)(y – y₀) + b( y – y₀)² = 0
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……(1) and l₂x + m₂ y = 0 …….. (2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of line parallel to (1) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(3)
The equation of line parallel to (2) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(4)

The combined equation of (3), (4) is
[l₁( x – x₀) + m₁( y – y₀)][l₂( x – x₀) + m₂(y – y₀)] = 0
⇒ l₁l₂(x – x₀)² + (l₁m₂ + l₂m₁)(x – x₀)(y – y₀) + m₁m₂(y – y₀)² = 0
⇒ a( x – x₀)² + 2h( x – x₀)( y – y₀) + b( y – y₀)2² = 0

Theorem:
The equation to the pair of lines passing through the origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0 .
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of the line perpendicular to (1) and passing through the origin is m₁x – l₁y = 0 ……….(3)
The equation of the line perpendicular to (2) and passing through the origin is m₂ x – l₂ y = 0 — (4)
The combined equation of (3) and (4) is
⇒ (m₁x – l₁ y)(m₂ x – l₂ y) = 0
⇒ m₁m₂x – (l₁m₂ + l₂m₁ )ny + l₁l₂ y = 0
bx² – 2hxy + ay² = 0

Theorem:
The equation to the lines passing through (x₀, y₀) and Perpendicular to ax² + 2hxy + by² = 0 is b(x – x₀)² – 2h(x – x₀)(y – y₀) + a(y – y₀)² = 0

Area of the triangle:
Theorem:
The area of triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h \ell m+b \ell^{2}\right|}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The given straight line is lx + my + n = 0 ………(3)
Clearly (1) and (2) intersect at the origin.
Let A be the point of intersection of (1) and (3). Then

Theorem:
The product of the perpendiculars from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 — (1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The lengths of perpendiculars from (α, β) to
the line (1) is p = \(\frac{\left|l_{1} \alpha+m_{1} \beta\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}\)
and to the line (2) is q = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)

∴ The product of perpendiculars is

Pair of Lines-Second Degree General Equation:
Theorem:
If the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines then
(i) Δ ≡ abc + 2fgh – af² – bg² – ch² =0 and
(ii) h² ≥ ab, g² ≥ ac, f² ≥ bc
Proof:
Let the equation S = 0 represent the two lines l₁x + m₁ y + n₁ = 0 and l₂ x + m₂ y + n₂ = 0. Then
ax2 + 2hxy + by2 + 2 gx + 2 fy + c
≡ (l₁x + m₁y + n₁)(l₂ x + m₂ y + n₂) = 0

Equating the co-efficients of like terms, we get
l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b, and l₁n₂ + l₂n₁ = 2g , m₁n₂ + m₂n₁ = 2 f , n₁n₂ = c

(i) Consider the product(2h)(2g)(2f)
= (l₁m₂ + l₂m₁)(l₁n₂ + l₂n₁)(m₁n₂ + m₂n₁)
= l₁l₂ (m₁²n₂ + m₂²n₁²) + m₁m₂ (l₁²n₂² + l₂²n₁²) + n₁n₂ (l₁²m₂² + l₂²m₁²) + 2l₁l₂m₁m₂n₁n₂
= l₁l₂[(m₁n₂ + m₂n₁) – 2m₁m₂n₁n₂] + m₁m₂[(l₁n₂ + l₂n₁) – 2l₁l₂n₁n₂] + n₁n₂[(l₁m₂ + l₂m₁) – 2l₁l₂m₁m₂] + 2l₁l₂m₁m₂n₁n₂
= a(4 f² – 2bc) + b(4g² – 2ac) + c(4h² – 2ab)
8 fgh = 4[af² + bg² + ch² – abc]
abc + 2 fgh – af² – bg² – ch² = 0

(ii) h² – ab = \(\left(\frac{l_{1} m_{2}+l_{2} m_{1}}{2}\right)^{2}\) – l1l2m1m2 = \(\frac{\left(l_{1} m_{2}+l_{2} m_{1}\right)^{2}-4-l_{1} l_{2} m_{1} m_{2}}{4}\)
= \(\frac{\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}{4}\) ≥ 0
Similarly we can prove g² > ac and f² ≥ bc

Note :
If A = abc + 2 fgh – af² – bg² – ch² = 0 , h² ≥ ab, g² ≥ ac and f² ≥ bc, then the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines

Conditions For Parallel Lines-Distance Between Them:
Theorem:
If S = ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of parallel lines then h² = ab and bg² = af². Also the distance between the two parallel lines is 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) (or) 2\(\sqrt{\frac{f^{2}-b c}{b(a+b)}}\)
Proof:
Let the parallel lines represented by S = 0 be
lx + my + n₁ = 0 ……….(1) lx + my + n₂ = 0 ………..(2)
ax² + 2hxy + 2gx + 2 fy + c
= (lx + my + n₁)(lx + my + n₂)

Equating the like terms
l² = a ………(3)
2lm = 2h …………(4)
m² = b ………..(5)
l(n₁ + n₂) = 2g …….(6)
m(n₁ + n₂) = 2 f ….(7)
n₁n₂ = c …….(8)

From (3) and (5), l²m² = ab and from (4) h² = ab .

Point of Intersection of Pair of Lines:
Theorem:
The point of intersection of the pair of lines represented by
a² + 2hxy + by² + 2gx + 2fy + c = 0 when h² > ab is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
Proof:
Let the point of intersection of the given pair of lines be (x₁, y₁).
Transfer the origin to (x₁, y₁) without changing the direction of the axes.
Let (X, Y) represent the new coordinates of (x, y). Then x = X + x₁ and y = Y + y₁.
Now the given equation referred to new axes will be
a( X + x₁)2 + 2h( X + x₁)(Y + y₁) +b(Y + y₁)2 + 2 g (X + x₁) + 2 f (Y + y₁) + c = 0
⇒ aX² + 2hXY + bY² + 2 X (ax₁ + hy₁ + g) + 2Y(hx₁ + by₁ + f) +(ax₁² + 2hx₁y₁ + by² + 2 gx₁ + 2 fy₁ + c) = 0

Since this equation represents a pair of lines passing through the origin it should be a homogeneous second degree equation in X and Y. Hence the first degree terms and the constant term must be zero. Therefore,
ax₁ + hy₁ + g = 0
hx₁ + by₁ + f = 0
ax₁² + 2hx₁ y₁ + by₁² + 2 gx₁ + 2 fyx + c = 0
But (3) can be rearranged as
x₁(ax₁ + hy + g) + y (hx₁ + byx + f) + (gx₁ + fq + c) = 0
⇒ gx₁ + fy₁ + c = 0 ………..(4)

Solving (1) and (2) for x₁ and y₁

Hence the point of intersection of the given pair of lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Theorem:
If the pair of lines ax² + 2hxy + by² = 0 and the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus then (a – b) fg+h(f² – g²) = 0.
Proof:
The pair of lines ax² + 2hxy + by² = 0 …………(1) is parallel to the lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 ……….. (2)

Now the equation
ax² + 2hxy + by² + 2gx + 2 fy + c + λ(ax² + 2hxy + by²) = 0

Represents a curve passing through the points of intersection of (1) and (2).
Substituting λ = -1, in (3) we obtain 2gx + 2fy + c = 0 …(4)
Equation (4) is a straight line passing through A and B and it is the diagonal \(\overline{A B}\)

The point of intersection of (2) is C = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
⇒ Slope of \(\overline{O C}=\frac{g h-a f}{h f-b g}\)
In a rhombus the diagonals are perpendicular ⇒ (Slope of \(\overline{O C}\)) (Slope of \(\overline{A B}\)) = -1
⇒ \(\left(\frac{g h-a f}{h f-b g}\right)\left(-\frac{g}{f}\right)\) = -1
⇒ g²h – afg = hf² – bfg
⇒ (a – b)fg + h(f² – g²) = 0
\(\frac{g^{2}-f^{2}}{a-b}=\frac{f g}{h}\)

Theorem:
If ax² + 2hxy + by² = 0 be two sides of a parallelogram and px + qy = 1 is one diagonal, then the other diagonal is y(bp – hq) = x(aq – hp)
proof:
Let P(x₁, y₁) and Q(x₂, y₂) be the points where the digonal

px + qy = 1 meets the pair of lines.
\(\overline{O R}\) and \(\overline{P Q}\) biset each other at M(α, β)
∴ α = \(\frac{x_{1}+x_{2}}{2}\) and β = \(\frac{y_{1}+y_{2}}{2}\)

Eliminating y from ax² + 2hxy+by² = 0
and px + qy = 1 ………..(2)
ax² + 2hx\(\left(\frac{1-p x}{q}\right)\) + b\(\left(\frac{1-p x}{q}\right)^{2}\) = 0
⇒ x² (aq² – 2hpq + bp²) + 2 x(hp – bp) + b = 0

The roots of this quadratic equation are x₁ and x₂ where
x₁ + x₂ = \(-\frac{2(h q-b p)}{a q^{2}-2 h p q-b p^{2}}\)
⇒ α = \(\frac{(b p-h q)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)

Similarly by eliminating x from (1) and (2) a quadratic equation in y is obtained and y₁,
y₂ are its roots where
y₁ + y₂ = \(-\frac{2(h p-a q)}{a q^{2}-2 h p q-n p^{2}}\) ⇒ β = \(\frac{(a q-h p)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)
Now the equation to the join of O(0, 0) and M(α, β) is (y – 0)(0 – α) = (x – 0)(0 – β)
⇒ αy = βx
Substituting the values of α and β, the equation of the diagonal OR
is y(bp – hq) = x(aq – hp).

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 3 The Straight Line to solve questions creatively.

Intermediate 1st Year Maths 1B The Straight Line Formulas

→ Equation of a horizontal line is y = k

→ Equation of a vertical line is x = h

→ If ‘θ ‘ is the inclination of a line then slope = m = tan θ

→ If m₁ m₂ are slopes of two lines and ‘θ ‘ is angle between them then tan θ = \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)

→ If two lines are parallel then their slopes are equal i.e., m₁ = m₂

→ If two lines are perpendicular, then m₁m₂ = -1

→ Equation of a line passing through (x₁, y₁) with slope ‘m’ is y – y₁ = m(x – x₁)

→ Equation of a line passing through origin with slope m is y = mx

→ Equation of the line passing through (x₁, y₁) and (x₂, y₂) is \(\) (or) (x – x₁)(y₁ – y₂) = (y – y₁)(x₁ – x₂)

→ Equation of the line with slope’m’ and having

• y-intercept ‘c’ is y = mx + c (slope intercept form)
• x-intercept ‘a’ is y = m(x – a)

→ Equation of a line in intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1

• Intersecting point on X – axis = (a, 0)
• Intersecting point on Y – axis = (0, b)

→ Area of Δ^le formed by the line with co-ordinate axis is = \(\frac{1}{2}\)|ab|

→ Area of Δ^le formed by the line ax + by + c = 0 with co-ordinate axis is \(\frac{c^{2}}{2|a b|}\)

→ If a point (x₁, y₁) divides the line segment between the co-ordinate axes in the ratio l: m then equation of the line is \(\frac{m x}{x_{1}}+\frac{l y}{y_{1}}\) = l + m

→ Equation of a line in normal form or perpendicular form is x cos α + y sin α = p where ‘α’ is the angle made by the perpendicular with +ve X – axis. ‘p’ is length of normal form origin to the line.

→ Perpendicular distance from (x₁, y₁) to the line ax + by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

→ Perpendicular distance from origin to the line ax + by + c = 0 is \(\frac{|c|}{\sqrt{a^{2}+b^{2}}}\)

→ The ratio in which the line L ax + by + c = 0 (ab ≠ 0) divides the line segment AB joining the points A(x₁, y₁) and B(x₂, y₂) is –\(\left(\frac{a x_{1}+b y_{1}+c}{a x_{2}+b y_{2}+c}\right)\) or –\(\frac{L_{11}}{L_{22}}\)
where L₁₁ = a₁x + b₁y + c; L₂₂ = a₂x + b₂y + c

→ If L₁₁, L₂₂ are having same sign or opposite sign’s then those points lies same side or opposite sides of the line L = 0 respectively.

→ If the foot of the perpendicular from (x₁, y₁) to the line ax + by + c = 0 is (α, β) then \(\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)

→ If the image of(x₁, y₁) w.r.t. the line ax +by + c = 0 is (α, β) then
\(\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{-2\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)

→ The point of intersection of two lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 is
\(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}} \cdot \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

→ If θ |0 ≤ θ ≤ π| is angle between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 then cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)
sin θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)
and tan θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\)

• The lines are perpendicular ⇔ a₁a₂ + b₁b₂ = 0
• Then lines are parallel ⇔ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\)

→ The equation of a line passing through the point (x₁, y₁) and parallel to the line ax + by + c = 0 is a(x – x₁) + b(y – y₁) = 0

→ The equation of a line passing through the point (x₁, y₁) and perpendicular to ax + by + c = 0 is b(x – x₁) – a(y – y₁) = 0

→ The distance between the two parallel straight lines a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₂ = 0 is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\)

→ Area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0,a ₂x + b₂y + d₁= 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is \(\left|\frac{\left(d_{1}-c_{1}\right)\left(d_{2}-c_{2}\right)}{a_{1} b_{2}-a_{2} b_{1}}\right|\)

→ Equation of the line parallel to X – axis and passing through (x₁, y₁) is y = y₁

→ Equation of the line parallel to Y- axis and passing through (x₁, y₁) is x = x₁

Concurrent lines-properties related to a Triangle
Theorem:
The medians of a triangle are concurrent.
Proof:
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of the triangle

Let D,E,F be the mid points of BC, CA, AB respectively
∴ D = \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)
E = \(\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)\)
F = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

Slope of \(\overline{A D}\) is \(\frac{\frac{y_{2}+y_{3}}{2}-y_{1}}{\frac{x_{2}+x_{3}}{2}-x_{1}}=\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\)

Equation of \(\overline{A D}\) is
y – y₁ = \(\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\)(x – x₁)
⇒ (y – y₁)(x₂ + x₃ – 2x₁) = (x – x₁)(y₂ + y₃ – 2y₁)
⇒ L₁ = (x – x₁)(y₂ + y₃ – 2y₁) – (y – y₁)(x₂ + x₃ – 2x₁) = 0

Similarly, the equations to \(\overline{B E}\) and \(\overline{C F}\) respectively are L₂ s (x – x₂)(y₃ + y₁ – 2y₂) – (y – y₂) (x₃ + x₁ – 2x₂) = 0.
L₃ = (x – x₃)(y₁ + y₂ – 2y₃) – (y – y₃) (x₁ + x₂ – 2x₃) = 0.
Now 1. L₁ + 1.L₂ + 1. L₃ = 0
The medians L₁ = 0, L₂ = 0, L₃ = 0 are concurrent.

Theorem:
The altitudes of a triangle are concurrent.
Proof:
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of the triangle ABC. Let AD, BE,CF be the altitudes.
Slope of \(\overline{B C}\) is \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\) and AD ⊥ BC

Slope of the altitude through A is – \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)

Equation of the altitude through A is y – y₁ = \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\) (x – x₁)
(y – y₁) (y₃ – y₂) = -(x – x₁) (x₃ – x₂)
L₁ = (x – x₁)(x – x₃) + (y – y₁)(y – y₃) = 0.

Similarly equations of the altitudes through B,C are
L₂ = (x – x₂) (x₃ – x₁) + (y – y₂) (y₂ – y₃) = 0,
L₃ = (x – x₃) (x₁ – x₂) + (y – y₃) (y₁ – y₂) = 0.
Now 1.L₁ + 1.L₂ + 1.L₃ = 0
The altitudes L₁ = 0, L₂ = 0, L₃ = 0 are concurrent.

Theorem:
The internal bisectors of the angles of a triangle are concurrent.

Theorem:
The perpendicular bisectors of the sides of a triangle are concurrent

Point of Intersection of Two Straight Lines:
Theorem:
The point of intersection of the two non parallel lines
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 is \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)
Proof:
The lines are not parallel ^ Slopes are not equal
⇒ \(\frac{-a_{1}}{b_{1}} \neq \frac{-a_{2}}{b_{2}}\) ⇒ a₁b₂ ≠ a₂b₁ ⇒ a₁b₂ – a₂b₁ ≠ 0
Let P (α, β) be the point of intersection. Then a₁ α + b₁β + C₁ = 0 and a₂ α + b₂β + c₂ = 0.
Solving by the method of cross multiplication

Point of intersection is P = \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

Theorem:
The ratio in which the line L = ax + by + c = 0 divides the line segment joining A(x₁, y₁), B(x₂, y₂) is -L₁₁ : L₂₂, where L₁₁ = L(x₁, y₁) = ax₁ + by₁ + c and L₂₂ = L(x₂, y₂) = ax₂ + by₂ + c.
Proof:
Let k : 1 be the ratio in which the line divides the line segment.

The point which divides in the ratio k : 1 is P = \(\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)\)

Since P lies on the line ax + by + c = 0 ⇒ a\(\left(\frac{k x_{2}+x_{1}}{k+1}\right)\) + b\(\left(\frac{k y_{2}+y_{1}}{k+1}\right)\) + c = 0
⇒ a(kx₂ + x₁) + b(ky₂ + y₁) + c(k + 1) = 0
⇒ k(ax₂ + by₂ + c) = -(ax₁ + by₁ + c)
⇒ k = –\(\frac{\left(a x_{1}+b y_{1}+c\right)}{a x_{2}+b y_{2}+c}\)
Required Ratio = -(ax₁ + by₁ + c) : (ax₂ + by₂ + c)
= -L₁₁: L₂₂.

Note:
The points A,B lie in the same side or opposite side of the line L = 0 according as L₁₁, L₂₂ have the same sign or opposite signs.

• The points A,B are opposite sides of the line L = 0 iff L₁₁ and L₂₂ have opposite signs.
• The points A,B are same side of the line L = 0 iff L₁₁ and L₂₂ have same sign.

Concurrent Lines:
Three or more lines are said to be concurrent if they have a point in common. The common point is called the point of concurrence.

Condition for Concurrency of Three Straight Lines:
Theorem:
The condition that the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 to be concurrent is a₃(b₁c₂ – b₂c₁) + b₃(c₁a₂ – c₂a₁) + c₃(a₁b₂ – a₂b₁) = 0.
Proof:
Suppose the given lines are concurrent.
Let P(α, β) be the point of concurrence.
Then a₁ α + b₁ β + c₁ = 0 ………..(1)
a₂ α + b₂ β + c₂= 0 ………..(2)
a₃ α + b₃ β + c₃= 0 ………….(3)

By solving (1) and (2) we get
α = \(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\)
β = \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\)

Therefore P = \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

Substituting P in eq.(3), we get
a₃\(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\) + b₃\(\left(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\) + c₃ = 0
a₃(b₁c₂ – b₂c₁) + b₃(c₁a₂ – c₂a₁) + C3 (a₁b₂ – a₂b₁) = 0 which is required Condition.

Above condition can be written in a determinant form as \(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|\) = 0

Problem:
Find the Condition that the lines ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0 to be Concurrent.
Answer:
Let P be the point of ConCurrenCe
aα + h β + g = 0 …………..(1)
hα + b β + f = 0 ………….(2)
gα + f β + c = 0 …………(3)

solving (1) and (2) we get

Sub these values in eq.(3)
g\(\left(\frac{h g-b g}{a b-h^{2}}\right)\) + f\(\left(\frac{g h-a f}{a b-h^{2}}\right)\) + c = 0
⇒ g(hf – bg) + f(gh – af) + C(ab – h²) = 0
⇒ fgh – bg² + fgh – af² + abc – ch² = 0
⇒ abc + 2fgh – af² – bg² – ch² = 0
The Condition is abc + 2fgh – af² – bg² – ch² = 0

Family of Straight Lines – Concurrent Straight Lines:
Theorem:
Suppose L₁ = a₁x + b₁y + c₁ = 0, L₂ = a₂x + b₂y + c₂ = 0 are two intersection lines.
(i) If (λ₁, λ₂) ≠ (0,0) then λ₁L₁ + λ₂L₂ = 0represents a straight line passing through the point of intersection of L₁ = 0 and L₂ = 0.
(ii) The equation of any line passing through the point of intersection of
L₁ = 0, L₂ = 0 is of the form where (λ₁, λ₂) ≠ (0,0).
Note: If L₁ = 0, L₂ = 0 are two intersecting lines, then the equation of any line other than L₂ = 0 passing through their point of intersection can be taken as L₁ + λL₂ = 0 where λ is a parameter.

Theorem:
If there exists three constants p,q,r not all zero such that
p(a₁x + b₁y + c₁) + q(a₂x + b₂y + c₂) + r(a₃x + b₃y + c₃) = 0 for all x and y then the three lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 in which no two of them are parallel, are concurrent.

Theorem:
Let L₁ = a₁x + b₁y + c₁ = 0, L₂ = a₂x + b₂y + c₂ = 0 represent two parallel lines.Then the straight line represented by λ₁L₁ + λ₂L₂ = 0 is parallel to each of the straight line L₁ = 0 and L₂ = 0.

A Sufficient Condition for Concurrency of Three Straight Lines:
Theorem:
If L₁ = a₁x + b₁y + c₁ = 0, L₂ = a₂x + b₂y + c₂ = 0, L₃ = a₃x + b₃y + c₃ = 0 are three straight lines, no two of which are parallel, and
if non-zero real numbers λ₁, λ₂, λ₃ exist such that λ₁ L₁ + λ₂ L₂ + λ₃ L₃ = 0 then the straight lines L₁ = 0, L₂ = 0 and L₃ = 0are concurrent.

Length of The Perpendicular From A Point to A Straight Line and Distance Between Two Parallel Lines
Theorem:
The perpendicular distance from a point P(x₁, y₁) to the line ax + by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)
Proof:

Let the axes be translated to the point P(x₁, y₁).
Let (X,Y) be the new coordinates of (x, y). Then x = X + x₁, y = Y + y₁
The transformed equation of the given line is
a(X + x₁) + b(Y + y₁) + c = 0
⇒ aX + bY + (ax₁ + by₁ + c) = 0
The perpendicular distance from the new origin P to the line is (from normal form) The perpendicular distance from a point
P(x₁, y₁) to the line ax+ by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

Distance Between Parallel Lines Theorem:
The distance between the two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)
Proof:
Given lines are ax + by + c₁ = 0 …………(1)
ax + by + c₂ = 0 …………(2)
Let P(x₁, y₁) be a point on the line (2).

Then
ax₁ + by₁ + c₂ = 0
ax₁ + by₁ = -c₂.
Distance between the parallel lines = Perpendicular distance from P to line (i)
= \(\frac{\left|a x_{1}+b y_{1}+c_{1}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)

Foot of The Perpendicular:
Theorem:
If (h, k) is the foot of the perpendicular from (x₁, y₁) to the line ax + by + c = 0 (a0, b0) then \(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\).
Proof:

Let A = (x₁, y₁) P = (h, k)
P lies on ax + by + c = 0
ah + bk + c = 0
ah + bk = – c
Slope of \(\overline{A P}\) is \(\frac{k-y_{1}}{h-x_{1}}\)
Slope of given line is \(-\frac{a}{b}\)

\(\overline{A P}\) is perpendicular to the given line
⇒ \(\left(\frac{k-y_{1}}{h-x_{1}}\right)\left(-\frac{a}{b}\right)\) = -1
⇒ \(\frac{k-y_{1}}{b}=\frac{h-x_{1}}{a}\)

By the law of multipliers in ratio and proportion

Image of A Point
Theorem:
If (h, k) is the image of (x₁, y₁) w.r.t the line ax + by + c = 0 (a ≠ 0, b ≠ 0), then \(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}=\frac{-2\left(x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\).
Proof:
Let A(x₁, y₁), B(h, k)
Mid Point of is P = \(\left(\frac{x_{1}+h}{2}, \frac{y_{1}+k}{2}\right)\)
Since B is the image of A,therefore mid point P lies on ax + by + c = 0.
a\(\left(\frac{x_{1}+h}{2}\right)\) + b\(\left(\frac{y_{1}+k}{2}\right)\) + c = 0
⇒ ax₁ + by₁ + ah + bk + 2c = 0
⇒ ah + bk = -ax₁ + by₁ – 2c.

Slope of \(\overline{A B}\) is \(\frac{k-y_{1}}{h-x_{1}}\)
Slope of given line is \(-\frac{a}{b}\)
\(\overline{A B}\) is perpendicular to the given line a
⇒ \(\left(\frac{k-y_{1}}{h-x_{1}}\right)\left(-\frac{a}{b}\right)\) = -1
⇒ \(\frac{k-y_{1}}{b}=\frac{h-x_{1}}{a}\)

By the law of multipliers in ratio and proportion

Note :

• The image of (x₁, y₁) w.r.t the line x = y is (y₁, x₁)
• The image of (x₁, y₁) w.r.t the line x + y = 0 is (-y₁, -x₁)

Theorem:
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. Then the area of the parallelogram so formed is
\(\left|\frac{(p-q)(r-s)}{b c-a d}\right|\)
Proof:
Let L₁ = ax + by + p = 0
L₂ = ax + by + q = 0
L₃ = cx + dy + r = 0
L₄ = cx + dy + s = 0
Clearly
L₁ ∥ L₂ and L₃ ∥ L₄. So L₁ and L₃ are nonparallel. Let be the angle between L₁ and L₃.
Let d₁ = distance between L₁ and L₂= \(\frac{|p-q|}{\sqrt{a^{2}+b^{2}}}\)
Let d₂ = distance between L₃ and L₄ = \(\frac{|r-s|}{\sqrt{c^{2}+d^{2}}}\)
Now cos θ = \(\frac{|a c+b d|}{\sqrt{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\) and sin θ = \(\sqrt{\frac{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)-(a c+b d)^{2}}{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\)
= \(\frac{|b c-a d|}{\sqrt{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\)
Now area of the parallelogram is \(\frac{d_{1} d_{2}}{\sin \theta}=\left|\frac{(p-q))(r-s)}{b c-a d}\right|\)

Angle Between Two Lines
Theorem:
If θ is an angle between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 then
cos θ = ±\(\frac{a_{1} a_{2}+b_{1} b_{2}}{\sqrt{a_{1}^{2}+b_{2}^{3}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)
Proof:
The lines passing through the origin and parallel to the given lines are
a₁x + b₁y = 0, ……… (1)
a₂x + b₂y = 0. …………….(2)

Let θ₁, θ₂ be the inclinations of (1) and (2) respectively (θ1 > θ2)
Now θ is an angle between (1) and (2)
θ = θ₁ – θ₂
P(-b₁, a₁) satisfies eq(1), the point lies on (1)
Similarly, Q(-b₂, a₂) lies on (2)

Let L and M be the projection of P, Q respectively on the x – axis.

Note :

• If is the acute angle between the lines then cos θ = \(\frac{\left|a_{1} a_{1}+b_{1} b_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)
• If is an angle between two lines, then is another angle between the lines.
• If is an angle between two lines are not a right angle then the angle between the lines means the acute angle between the lines.
• If 0 is an angle between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 then tan θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\)
• If is the acute angle between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 then
  tan θ = \(\left|\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\right|=\left|\frac{a_{1} / b_{1}-a_{2} / b_{2}}{\left(a_{1} a_{2}\right) /\left(b_{1} b_{2}\right)+1}\right|\)
  = \(\left|\frac{\left(-a_{1} / b_{1}\right)-\left(-a_{2} / b_{2}\right)}{1+\left(-a_{1} / b_{1}\right)\left(-a_{2} / b_{2}\right)}\right|\) where mj, m2 are the slopes of the lines.

Theorem:
The equation of the line parallel to ax + by + c = 0 and passing through (x₁, y₁) is a(x – x₁) + b(y – y₁) = 0.
Proof:
Slope of the given line is -a/b.
⇒ Slope of the required line is -a/b.(lines are parallel)
Equation of the required line is
y – y₁ = –\(\frac{a}{b}\)(x – x₁)
b(y – y₁) = -a(x – x₁)
a(x – x₁) + b(y – y₁) = 0.

Note :

• The equation of a line parallel to ax + by + c = 0 may be taken as ax + by + k = 0.
• The equation of a line parallel to ax + by + c = 0 and passing through the origin is ax + by = 0.

Theorem:
The equation of the line perpendicular to ax + by + c = 0 and passing through (x₁, y₁) is b(x – x₁) – a(y – y₁) = 0.
Proof:
Slope of the given line is -a/b. ⇒ Slope of the required line is b/a. (since product of slopes = -1)
Equation of the required line is y – y₁ = \(\frac{b}{a}\) (x – x₁)
a(y – y₁) = b(x – x₁)
b(x – x₁) – a(y – y₁) = 0.

Note :

• The equation of a line perpendicular to ax + by + c = 0 may be taken as bx – ay + k = 0
• The equation of a line perpendicular to ax + by + c = 0 and passing through the origin is bx – ay = 0.

Concurrent Lines- Properties Related to A Triangle
Theorem:
The medians of a triangle are concurrent.
Proof:
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of the triangle

Let D, E, F be the mid points of \(\overline{B C}, \overline{C A}, \overline{A B}\) respectively

Equation of \(\overline{A D}\)
y – y₁ = (x – x₁)
(y – y₁) (x₂ + x₃ – 2x₁) = (x – x₁)(y₂ + y₃ – 2y₁)
⇒ L₁ ≡ (x – x₁)(y₂ + y₃ – 2y₁) – (y – y₁) (x₂ + x₃ – 2x₁) = 0.

Similarly, the equations to \(\overline{B E}\) and \(\overline{C F}\) respectively are L₂ ≡ (x – x₂)(y₃ + y₁ – 2y₂) – (y – y₂) (x₃ + x₁ – 2x₂) = 0.
L₃ ≡ (x – x₃)(y₁ + y₂ – 2y₃) – (y – y₃) (x₁ + x₂ – 2x₃) = 0.
Now 1. L₁ + 1.L₂ + 1. L₃ = 0
The medians L₁ = 0, L₂ =0, L₃ = 0 are concurrent.

Theorem:
The altitudes of a triangle are concurrent.
Proof:
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of the triangle ABC.
Let AD, BE,CF be the altitudes.
Slope of \(\overline{B C}\) is \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\) and AD ⊥ BC
Slope of the altitude through A is \(-\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)

Equation of the altitude through A is y – y₁ = \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)(x – x₁)
(y – y₁) (y₃ – y₂) = -(x – x₁) (x₃ – x₂)
L₁ = (x – x₁)(x₂ – x₃) + (y – y₁)(y₂ – y₃) = 0.

Similarly equations of the altitudes through B,C are
L₂ = (x – x₂) (x₃ – x₁) + (y – y₂) (y₂ – y₃) = 0
L₃ = (x – x₃) (x₁ – x₂) + (y – y₃) (y₁ – y₂) = 0.
Now 1.L₁ + 1.L₂ + 1.L₃ = 0
The altitudes L₁ = 0, L₂ =0, L₃ = 0 are concurrent.

Theorem:
The internal bisectors of the angles of a triangle are concurrent.

Theorem:
The perpendicular bisectors of the sides of a triangle are concurrent

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 2 Transformation of Axes to solve questions creatively.

Intermediate 1st Year Maths 1B Transformation of Axes Formulas

→ x = x’ + h ⇒ x’ = x – h

→ y = y’ + k ⇒ y’ = y – k

→ To make the first degree term absent, origin should be shifted to \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If the equation is ax² + by² + 2gx + 2fy + c = 0, origin should be shifted to be \(\left(-\frac{g}{a},-\frac{f}{b}\right)\)

x = x’ cos θ – y’ sin θ
x’ = x cos θ + y sin θ
y = x’ sin θ + y’ cos θ
y’ = – x sin θ + y cos θ

→ To make the xy term absent, axes should be rotated through an angle θ given by
tan 2θ = \(\frac{2 h}{a-b}\)

Rotation Of Axes (Change Of Direction):
I. Definition: If thc axes are rotated through an angle in the same plane by keeping the origin constant, then the transformation is called Rotation of axes.

2. Theorem: To find the co-ordinates of a point (x,y)are transformed to (X, Y)when the axes are rotated through an angle ‘6’ about the origin in the same plane.
Proof: Let x’Ox, yOY’ are the original axes
Let P(x,y)be the co-ordinates of the point in the above axes.
After rotating the axes through an angle ‘θ’, then the co-ordinates of P be (X, Y) w.r.t the new axes X’OX and YOY’ as in figure.

Since θ is the angle of rotation, then ∠xOA = ∠yOY = θ as in the figure.
Since L, M is projections of P on Ox and OX respectively. We can see that ∠LPM = ∠xOA = θ
Let N be the projection to PL from M
Now x = OL = OQ – LQ = OQ – NM
= OM cos θ – PM sin θ
= A cos θ – Y sin θ
y = PL = PN + NL = PN + MQ
PM cos θ + OM sin θ = Ycos θ + X sin θ
x = X cos θ – Y sin θ and
y = Y cos θ + X sin θ ………….(1)
Solving the above equations to get X and Y, then X = x cos θ + y sin θ and Y = -x sin θ + y cos θ ………… (2)

From (1) and (2) we can tabulate

Note:

• If the axes are turned through an angle ‘ θ ’, then the equation of a curve f (x, y) = 0 is transformed to f (X cos θ – Y sin θ, X sin θ + Y cos θ) = 0
• If f (X, Y) = 0is the transformed equation of a curve when the axes are rotated through an angle ‘ θ ’ then the original equation of the curve is f (x cos θ + y sin θ, – x sin θ + y cos θ) = 0

Theorem:
To find the angle of rotation of the axes to eliminate xy term in the equation ax² + 2hxy + by² + 2 gx + 2 fy + c = 0
Proof: given equation is ax²+ 2hxy + by² + 2 gx + 2 fy + c = 0
Since the axes are rotated through an angle θ, then x = X cos θ – Y sin θ, y = X sin θ + Y cos θ

Now the transformed equation is
a (X cos θ – Y sinθ)² + 2h (X cos θ – Y sin θ)(X sin θ + Y cos θ) + b (X sin θ + Y cos θ)² + 2g (X cos θ – Y sin θ) + 2 f (X sin θ + Y cos θ) + c = 0

⇒ X²cos²θ + Y²sin²θ – 2XYcosθsinθ) + 2h [X²cos θ + XY(cos²θ – sin²θ) – Y sin²θ cos²θ] + b[X² sin²θ + Y² cos²θ + 2XY cos θ sin θ) + 2g (X cos θ – Y sin θ) + 2 f (X sin θ + Y cos θ) + c = 0

Since XY term is to be eliminated, coefficient of XY = 0.
⇒ 2 (b – a) cos θ sin θ + 2h (cos2 θ – sin2 θ) = 0
⇒ h cos 2θ + (b – a) sin 2θ = 0
⇒ 2hcos2θ = (a -b)sin 2θ
⇒ Angle of rotation (θ) = \(\frac{1}{2}\) Tan^-1\(\left(\frac{2 h}{a-b}\right)\)
Note: The angle of rotation of the axes to eliminate xy term in
ax² + 2hxy + ay² + 2 gx + 2 fy + c = 0 is \(\frac{\pi}{4}\)

Translation of Axes:
1. Definition: Without changing the direction of the axes, the transformation in which the origin is shifted to another position or point is called translation of axes.

2. Theorem: To find the co-ordinates of a point (x, y) are translated by shifting the origin to a point (x, y)
Proof :

Let xox¹ xox¹, yoy¹ be the original axes and A(x₁, y₁) be a point to which the origin is shifted
Let AX, AY be the new axes which are parallel to the original axes as in figure.
Let P be any point in the plane whose coordinates w.r.t old system are (x, y)
And w.r.t new axes are (X, Y) .
From figure, A = (x₁, y₁) then AL = y₁, OL = x₁,
P(x, y )then x = ON = OL + LN = OL + AM = x₁ + X = X + x₁
Hence x = X + x₁.
y = PN = PM +MN = X+ AL = X + y₁
therefore, y = Y + y₁
hence x = X + x₁ y = Y + y₁

3. Theorem: To find the point to which the origin is to be shifted by translation of axes to eliminate x, y terms(first degree terms) in the equation ax² + 2xhy + by² + 2gx + 2fy + c = 0 (2 = ab)
Proof : given equation is ax2 + 2 xhy + by2 + 2gx + 2fy + c = 0
Let (x1, y1) be a point to which the origin is shifted by translation Let (X ,Y) be the new co-ordinates of the point (x, y) .
the equations of the transformation are x = X + x₁, y = Y + y₁
Now the transformed equation is
a (X + x₁)² + 2h (X + x₁)(Y + y₁)+ b (Y + y₁)² + 2g (X + x₁) + 2f [Y + y₁] + c = 0
⇒ a(X² + 2x₁X + x₁²) + 2h(XY + x₁Y + y₁X + x₁y₁) + b(Y² + 2y₁Y + y₁²) + 2g(X + x₁) + 2f (Y + y₁) + c = 0
⇒ aX² + 2hXY + bY² + 2X [ax₁ + hy₁ + g) + 2Y(hx₁ + by₁ + f ) + (ax₁² + 2x₁ + by₁² + 2gx₁ + 2fy₁ + c) = 0

Since x, y terms (the first degree terms) are to be eliminated
ax₁ + hy₁ + g = 0
hx₁ + by₁ + f = 0
Solving these two equations by the method of cross pollination

\(\frac{x_{1}}{h f-b g}=\frac{y_{1}}{g h-a f}=\frac{1}{a b-h^{2}}\) ⇒ x₁ = \(\frac{h f-b g}{a b-h^{2}}\) y₁ = \(\frac{g h-a f}{a b-h^{2}}\)
New Origin is ⇒ (x₁, y₁) = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Note:
(i) The point to which the origin has to be shifted to eliminate x, y terms by translation in the equation ax2 + by2 + 2gx + 2 fy + c = 0is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)
If b = a, then the new origin is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

(ii) The point to which the origin has to be shifted to eliminate x, y terms by translation of axes in the equation a(x + x₁)² + b(y + y₁)² = c is (-x₁, – y₁)

(iii) The point to which the origin has to be shifted to eliminate x, y terms by translation in the equation 2hxy + 2gx + 2 fy + c = 0 is \(\left(\frac{-f}{h}, \frac{-g}{h}\right)\)
l h ’ h I
Theorem : To find the condition that the equation ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 to be in the form aX² + 2hXY + bY² = 0 when the axes are translated. (h² ≠ ab)
Proof : From theorem 3, we get
ax₁ + hy₁ + g = 0 ………..(1)
hx₁ + by₁ + f = 0 …………(2)

Solving (1) and (2),
(x₁, y₁) = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g f-a f}{a b-h^{2}}\right)\)

Since the equation is to be in the form of aX² + 2hXY + bY² = 0 ,then for this we should have axj² + 2hx₁y₁ + by² + 2 gx₁ + 2 fy₁ + c = 0
⇒ (ax₁ + hyx + g) x₁ + (x₁ + byx + f) yx + (gx + fy₁ + c ) = 0
⇒ (0).x + (0), y + (gx₁ + fyx + c) = 0 from(1) and (2) ⇒ gx₁ + fyx + c = 0 …………..(3)

Substituting x₁, y₁ in (3), we get
g\(\left(\frac{h f-b g}{a b-h^{2}}\right)\) + f\(\left(\frac{g h-a f}{a b-h^{2}}\right)\) + c = 0
⇒ fgh – bg² + fgh – af² + abc + -ch² = 0
⇒ abc + 2 fgh – af² – bg² – ch² = 0

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 1 Locus to solve questions creatively.

Intermediate 1st Year Maths 1B Locus Formulas

→ PQ = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)

→ OP = \(\sqrt{x_{1}^{2}+y_{1}^{2}}\)

→ P divides A(x₁, y₁) and B(x₂, y₂) in the ratio m : n

→ Co-ordinates of P are \(\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)

→ Midpoint = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

→ ΔABC = \(\frac{1}{2}\)[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃(y₁ – y₂)] = \(\frac{1}{2}\)\(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)

→ Area of the Quadrilateral = \(\frac{1}{2}\)|x₁(y₂ – y₄) + x₂(y₃ – y₁) + x₃(y₄ – y₂) + x₄(y₁ – y₃)|

→ Centroid G = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

→ Incentre I = \(=\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\)

→ Ex-centre I₁ = \(\left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)\)

Distance Between two points:

• The distance between two points A(x₁, y₁) and B(x₂, y₂)
  AB (orBA) = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\)
• The distance from origin O to the point A(x₁, y₁) OA = \(\sqrt{x_{1}^{2}+y_{1}^{2}}\)
• The distance between two points A(x₁, 0) and B(x₂, 0) lying on the X – axis is AB= \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+(0-0)^{2}}=\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = |x₁ – x₂|
• The distance between two points C(0, y₁) and D(0, y₂) lying on the Y-axis is CD = |y₁ – y₂|

Section Formula:

• The point P which divides the line segment joining the points A(x₁, y₁). B(x₂, y₂) in the ratio m : n internally is given by P = \(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\)
• If P divides in the ratio m:n externally then P = \(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}\) (m’ n)

Note: If the ratio m : n is positive then P divides internally and if the ratio is negative P divides externally.

Mid Point:
The mid point of the line segment joining A(x₁, y₁) and B(x₂, y₂) is \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

Points Of Trisection:
The points which divide the line segment \(\overline{A B}\) in the ratio 1: 2 and 2: 1 (internally) are called the points of trisection of \(\overline{A B}\)

Area Of A Triangle:
The area of the triangle formed by the points A(x₁, y₁), B(x₂, y₂) C(x₃, y₃) is
Area = \(\frac{1}{2}\)|(x₁y₂ – x₂y₁) +(x₂y₃ – x₃y₂) + (x₃y₁ – x₁y₃)|
i.e., Area of ABC = \(\frac{1}{2}\)|Σ(x₁y₂ – x₂y₁)|

Note:

• The area of the triangle formed by the points (x₁, y₁)(x₂, y₂).(x₃, y₃) is the positive value of the determinant \(\frac{1}{2}\)\(\left|\begin{array}{ll}
  x_{1}-x_{2} & y_{1}-y_{2} \\
  x_{2}-x_{3} & y_{2}-y_{3}
  \end{array}\right|\)
• The area of the triangle formed by the points (x₁, y₁)(x₂, y₂) and the origin is \(\frac{1}{2}\)|x₁y₂ – x₂y₁|

Area Of A Quadrilateral:
The area of the quadrilateral formed by the points (x₁, y₁) (x₂, y₂), (x₃, y₃), (x₄, y₄) taken in that order is
\(\frac{1}{2}\)|x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₄ – x₄y₃ + x₄y₁ – x₁y₄|

Note:
The area of the quadrilateral formed by the points (x₁, y₁) (x₂, y₂) (x₃, y₃),(x₄, y₄) taken in order is
Area = \(\frac{1}{2}\left|\begin{array}{ll}
x_{1}-x_{3} & y_{1}-y_{3} \\
x_{2}-x_{4} & y_{2}-y_{4}
\end{array}\right|\)

Centres Of A Triangle:
Median: In a triangle, the line segment joining a vertex and the mid point of its opposite side is called a median of the triangle. The medians of a triangle are concurrent.
The point of concurrence of the medians of a triangle is called the centroid (or) centre of gravity of the triangle. It is denoted by G.

In Centre Of A Triangle:
Internal bisector : The line which bisects the internal angle of a triangle is called an internal angle bisector of the triangle.
The point of concurrence of internal bisectors of the angles of a triangle is called the incentre of the triangle, it is denoted by I.
I = \(\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\)

Ex-centres Of A Triangle:
The point of concurrence of internal bisector of angle A and external bisectors of angles B, C of ABC is called the ex-centre opposite to vertex A. It is denoted by I₁. The excentres of ABC opposite to the vertices B, C are respectively denoted by I₂, I₃.

• I₁ = Excentre opposite to A = \(\left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)\)
• I₂ = Excentre oppposite to B = \(\left(\frac{a x_{1}-b x_{2}+c x_{3}}{a-b+c}, \frac{a y_{1}-b y_{2}+c y_{3}}{a-b+c}\right)\)
• I₃ = Excentre opposite to C = \(\left(\frac{a x_{1}+b x_{2}-c x_{3}}{a+b-c}, \frac{a y_{1}+b y_{2}-c y_{3}}{a+b-c}\right)\)

Ortho Centre Of A Triangle:
Altitude: The line passing through vertex and perpendicular to opposite side of a triangle is called an altitude of the triangle. Altitudes of a triangle are concurrent. The point of concurrence is called the ortho centre of the triangle. It is denoted by “O” or H’.

Circum centre of a Triangle:
Perpendicular bisector: The line passing through mid point of a side and perpendicular to the side is called the perpendicular bisector of the side.
The perpendicular bisectors of the sides of a triangle are concurrent. The point of concurrence is called the circurn centre or the triangle. It is denoted by S.

→ We have mentioned that the Coordinate Geometry unifies the ideas of Algebra and Geometry. Now, in this chapter, we study a set of points, which satisfies certain geometric conditions that can also be represented in the form of algebraic equation.

Definition of Locus :
Let us call a set of geometric conditions ‘consistent’ if there is atleast one point satisfying that set of conditions. For example when A = (1, 0) and B = (3,0), the condition ‘the sum of distances of a point P from A and B is equal to 2’ is consistent, where as the condition ‘the sum of distances of a point Q from A and B is equal to 1’ is not consistent, because there is no point Q such that QA + QB = 1 (since AB = 2)
By the locus of a point we mean the set of all positions that it can take when it is subjected to certain consistent geometric conditions.
Or
Locus:

• The set of all points (and only those points) which satisfy the given geometrical condition(s) (or properties) is called a locus.
  Eg. The set of points in a plane which are at a constant distance r from a given point C is a locus. Here the locus a circle.
• The set of points in a plane which are equidistant from two given points A and B is a locus. Here the locus is a straight line and it is the perpendicular bisector of the line segment joining A and B.

Equation of Locus :
It is clear that, every point on the locus satisfies the given conditions and every point which satisfies the given conditions lies on the locus.
By the equation of a locus we mean an algebraic description of the locus. It is obtained by translating the geometric conditions satisfied by the points on the locus, into equivalent algebraic conditions.

Algebraic descriptions give rise to algebraic equations which sometimes contain more than what is required by the geometric conditions. Thus locus may be a part of the curve represented by the algebraic equation. Usually we call this algebraic equation as the equation of locus. However, to get the full description of the locus, the exact part of the curve, points of which satisfy the given geometric description, must be specified.
or
Equation of a Locus:
An equation f(x, y) = O is said to be the equation of a locus S if every point of S satisfies f(x, y) = O and every point that satisfies f(x, y) = O belongs to S.

An equation of a locus is an algebraic description of the locus. This can be obtained in the following way

• Consider a point P(x, y) on the locus
• Write the geometric condition(s) to bc satisfied by P in terms of an equation or in equation in symbols.
• Apply the proper formula of coordinate geometry and translate the geometric condition(s) into an algebraic equation.
• Simplify the equation so that it is free from radicals. The equation thus obtained is the required equation of locus.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(d)

I.

Question 1.
If y = \(\frac{2x+3}{4x+5}\) then find y”
Solution:
y = \(\frac{2x+3}{4x+5}\)
Differentiating w.r.to x

Question 2.
y = ae^nx + be^-nx nx then prove that y” = n²y
Solution:
y = ae^nx + be^-nx
y₁ = na e^nx – nb e^-nx
y_{2 = n² . ae^nx + n² be^-nx
y”= n² (ae^nx + b.e^-nx)
= n²y}

II.

Question 1.
Find the second order derivatives of the following functions f(x)
i) cos³ x
Solution:
y = cos³ x = \(\frac{1}{4}\) [cos 3x + 3 cos x]
y’ = \(\frac{1}{4}\) [- 3 sin 3x – 3 sin x]
y” = \(\frac{1}{4}\) (- 9 cos 3x – 3 cos x)
= – \(\frac{1}{4}\) (3 cos x + 9 cos 3x)
= – \(\frac{3}{4}\) (cos x + 3 cos 3x)

ii) y = sin⁴ x
Solution:
y = sin⁴ x = (sin²x)² = (\(\frac{(1-\cos 2 x)^{2}}{2}\))
= \(\frac{1}{4}\) [1 – 2 cos 2x + cos² 2x]
= \(\frac{1}{4}\) [1 – 2cos 2x + \(\frac{1+\cos 4 x}{2}\)]
= \(\frac{1}{8}\) [2 – 4 cos 2x + 1 + cos 4x]
= \(\frac{1}{8}\)(3 – 4 cos 2x + cos 4x)
y’ = \(\frac{1}{8}\) (8 sin 2x – 4 sin 4x)
y” = \(\frac{1}{8}\) (16 cos 2x-16 cos 4x)
= 2 (cos 2x – cos 4x)

iii) log (4x² – 9)
Solution:
y = log (4x² – 9)
= log (2x – 3) (2x + 3)
= log (2x – 3) + log (2x + 3)

iv) y = e^-2x sin³ x
Solution:
y = e^-2x. sin³ x
y’ = e^-2x (3 sin² x. cos x) + sin3 x (e^-2x) (-2)
= e^-2x [3 sin² x. cos x – 2 sin³ x)
\(\frac{d^{2} y}{d x^{2}}\) = e^-2x [3 sin² x (- sin x) + 3 cos x (2 sin x) cos x – 6 sin² x cos x) – 2. e^-2x
[3 sin² x. cos x – 2 sin³ x]
= e^-2x [6 sin x. cos² x – 6 sin² x. cos x – 3 sin³ x. – 6 sin² x. cos x + 4 sin³ x)
= e^-2x [sin³ x – 12 sin² x. cos x + 6 sin x. cos² x)

v) e^x sin x cos 2x
Solution:
y = e^x. sin x. cos 2x = \(\frac{e^{x}}{2}\) (2 cos 2x. sin x)
= \(\frac{e^{x}}{2}\) (sin 3x – sin x)
y’ = \(\frac{1}{2}\) [e^x(3 cos 3x – cos x) + e^x (sin 3x – sin x)
y”= \(\frac{1}{2}\) [ex (- 9 sin 3x + sin x) + e^x (3 cos 3x – cos x) + e^x (3 cos 3x – cos x) + e^x (sin 3x – sin x)]
= \(\frac{e^{x}}{2}\) [- 9 sin 3x + sin x + 3 cos 3x – cos x + 3 cos 3x cos x + sin 3x – sin x]
= \(\frac{e^{x}}{2}\) [6 cos 3x – 8 sin 3x – 2 cos x]
= e^x [3 cos 3x – 4 sin 3x – cos x]

vi) Tan^-1\(\frac{1+x}{1-x}\)
Solution:
y = tan^-1\(\frac{1+x}{1-x}\)
Put x = tan θ

∴ f(x) = \(\frac{\pi}{4}\) + tan^-1 (x)
Diff. w.r.to x
f'(x) = 0 + \(\frac{1}{1+x^{2}}\)
f”(x) = (-1) (1 + x²)^-2 (2x)
f”(x) = \(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\)

vii) tan^-1\(\frac{3x-x^{3}}{1-3x^{2}}\)
Solution:
f(x) = tan^-1\(\frac{3x-x^{3}}{1-3x^{2}}\) ; Put x = tan θ
f(x) = tan^-1\(\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)\)
= tan^-1(tan 3θ) = 3θ
∴ f(x) = tan^-1 (x)
f'(x) = 3\(\frac{1}{1+x^{2}}\) = \(\frac{3}{1+x^{2}}\)
Again Diff. w.r.to x
f”(x) = (3) (-1) (1 + x²)^-2 (2x)
⇒ f”(x) = \(\frac{-6 x}{\left(1+x^{2}\right)^{2}}\)

II. Prove the following.
If y = ax^{n + 1} + bx^-n
then x²y” = n(n + 1) y.
Solution:
y = ax^{n + 1} + bx^-n
y_{1 = (n + 1). axⁿ – n bx^-n-1
y2 = n(n + 1). ax^n-1 + n(n + 1) bx^-n-2
∴ x²y2 = n(n + 1) axⁿ⁺¹ + n(n + 1) bx^-n
= n(n + 1) (axn+1+bx_n) = n(n + 1) y
∴ x²y” = n(n + 1) y}

ii) If y = a cos x + (b + 2x) sin x, then y” + y = 4 cos x
Solution:
\(\frac{dy}{dx}\) = y’ = a(-sin x) + (b + 2x) \(\frac{d}{dx}\) (sin x) + sin x \(\frac{d}{dx}\) (b + 2x)
= – a sin x + (b + 2x) cos x + sin x.2
\(\frac{d^{2} y}{d x^{2}}\) = y” = – a cos x + (b + 2x) (- sin x) + cos x (2) + 2 cos x
L.H.S. = y” + y = – a cos x + (b + 2x) (- sin x) + 2 cos x + 2 cos x + a cos x+ (b + 2x) sinx = 4 cosx

iii) If y = 6 (x + 1) + (a + bx) e^3x then y” – 6y’ + 9y = 54 x + 18
Solution:
y’ = 6 \(\frac{d}{dx}\) (x + 1) + (a + bx) \(\frac{d}{dx}\) (e^3x) + e^3x \(\frac{d}{dx}\) (a + bx)
= 6(1) + (a + bx) 3e^3x + e^3x. b.
y” = 0 + 3 (a + bx) \(\frac{d}{dx}\) e^3x + 3e^3x \(\frac{d}{dx}\) (a + bx) + b \(\frac{d}{dx}\) (e^3x)
= 3(a + bx) 3e^3x + 3e^3x(b) + b. 3e^3x
= 9(a + bx) e^3x + 6b e^3x
Now L.H.S. = y” – 6y’ + 9y = 9(a + bx) e^3x + 6be^3x – 6[6 + 3(a + bx)e^3x + be^3x] + 9 [6(x + 1) + (a + bx)e^3x]
= 9(a + bx) e^3x + 6b e^3x – 36 – 18(a + bx)e^3x – 6be^3x + 54x + 54 + 9(a + bx) e^3x = 54x + 18

iv) If ay⁴ = (x + b)⁵ then 5y y” = (y’)²
Solution:
ay⁴ = (x + b)⁵ ; y⁴ = \(\frac{\left(x+b\right)^{2}}{a}\)

v) If y = a cos (sin x) + b sin (sin x) then y” + (tan x) y’ + y cos² x = 0
Solution:
y = a cos (sin x) + b sin (sin x) …………. (1)
Differentiating w.r.to x
y₁ = – a sin (sin x) cos x + b cos (sin x). cos x = [- a sin (sin x) + b cos (sin x)j cos x ………. (2)
Differentiating again w.r.to x.
y₂ = – sin x [-a sin (sin x) + b cos (sin x)] + cos x [- a cos (sin x) cos x – b sin (sin x) cos x]
= – sin x. \(\frac{y_{1}}{\cos x}\) = -cos² x.y
From (1) and (2),
⇒ y₂ + (tan x) y₁ + y cos² x = 0

III.

i) If y = 128 sin³ x cos⁴ x, then find y”.
Solution:
f(x) = 128 sin³ x. cos⁴ x
D.w.r. to x
f'(x)= 128 [sin³ x {4cos³ x (-sinx)} + cos⁴ x {3sin² x. cos x}]
= 128 [3sin² x cos⁵ x – 4sin⁴ x cos³ x]
Again D.w.r to x.
f”(x)= 128 {3 [sin² x 5cos⁴ x . (-sinx) + cos⁵ x. 2sinx cosx] – 4 [sin⁴ x. 3. cos2x (-sinx) + cos³ x. 4 sin³ x. cos x]}
= 128 [-15sin³ x . cos⁴ x + 6sin x cos⁶ x + 12 sin⁵ x cos² x- 16sin³ x cos⁴ x]
f”(x)= 128 [6sinx cos6x + 12sin⁵ x. cos² x – 31sin³ x. cos⁴ x]

ii) If y = sin 2x sin 3x sin 4x, then find y”.
Solution:
f(x) = sin 2x sin 3x sin 4x
= \(\frac{1}{2}\)sin 2x [2sin 4x. sin 3x]
= \(\frac{1}{2}\)sin 2x [cos x – cos 7x]
= \(\frac{1}{2}\) [sin 2x . cos x – cos 7x. sin 2x]
= \(\frac{1}{2}\)×\(\frac{1}{2}\) [2 sin2x. cosx – 2 cos7x. sin2x]
= \(\frac{1}{4}\) [(sin3x + sinx) – (sin9x – sin5x)]
= \(\frac{1}{4}\) [-sin9x + sin5x + sin3x + sinx]
D.w.r. to x
f'(x) = \(\frac{1}{4}\) [-9 cos 9x + 5 cos 5x + 3 cos 3x + cos x]
D.w.r. to x
f”(x) = \(\frac{1}{4}\) [81 sin 9x – 25sin 5x – 9sin 3x – sin x]

iii) If ax² + 2hxy + by² = 1 then prove that
\(\frac{d^{2} y}{d x^{2}}=\frac{h^{2}-a b}{(h x+b y)^{3}}\)
Solution:
Given ax² + 2hxy + by² = 1
Differentiating w.r. to x
a.2x + 2h(x. \(\frac{dy}{dx}\) + y) + b . 2y\(\frac{dy}{dx}\) = 0
2ax + 2hx. \(\frac{dy}{dx}\) + 2hy + 2by. \(\frac{dy}{dx}\) = 0
2(hx + by).\(\frac{dy}{dx}\) = -2(ax + hy)
\(\frac{dy}{dx}\) = \(\frac{-2(ax +hy)}{2(hx+by)}\) = \(\frac{(ax+hy)}{(hx+by)}\) ……… (1)
Differentiating again w.r. to x, \(\frac{d^{2} y}{d x^{2}}\)

iv) If y = ae^-bx cos(cx + d) then prove that y” + 2by’ + (b² + c²) y = 0.
Solution:
Given y = ae^-bx cos(cx + d) ……….. (1)
y₁ = a[e^-bx {- sin (cx + d)}. c. + cos (cx + d) e^-bx (-b)}
= – a. e^-bx [c sin (cx + d) + b cos (cx + d)]
= – ac.e^-bx sin (cx + d) – by
y₁ + by = – ac. e^-bx sin (cx + d) ………. (2)
Differentiating once w.r.to x
y₂ + by₁ = – ac(- b) e^-bx sin (cx + d) – ac e^-bx cos (cx + d) (+ c)
= abce^-bx sin (cx + d) -ac² e^-bx cos (cx + d)
= – b (y₁ + by) c²y [From (1) and (2)]
⇒ y₂ + by₁ + by₁ + b²y + c²y = 0.
⇒ y₂ + 2by₁ + (b² + c²) y = 0 (or)
y” + 2 by’ + (b² + c²) y = 0

v) If y = \(e^{\frac{-k}{2} x}\) (a cos nx + b sin nx) prove that then y” + ky’+ (n² + \(\frac{k^{2}}{4}\))y = 0
Solution:
Given y = \(e^{\frac{-k}{2} x}\) (a cos nx + b sin nx) ………… (1)
∴ y₁ = \(e^{\frac{-k}{2} x}\) (-n. a sin nx + n. b cos nx) + (a cos nx + b sin nx).\(e^{\frac{-k}{2} x}\) (-\(\frac{k}{2}\))
= –\(\frac{k}{2}\) . y – n.e^-kx/2 (a sin nx + b cos nx)
∴ y₁ + \(\frac{k}{2}\) y = – n.e^-kx/2 (a sin nx + b cos nx) …………. (2)
Differentiating w.r. to x k
y₂ + \(\frac{k}{2}\) y₁ = -n[{e^-kx/2 (- na cos nx – nb sin nx)}] + {(a sin nx +b cos nx).e^-kx/2 \(\frac{k}{2}\)}
= – n² e^-kx/2(a cos nx + b sin nx) – \(\frac{k}{2}\){-n.e^-kx/2 (a sin nx + b cos nx)}

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(c)

I.

Question 1.
Find the derivatives of the following functions.
i) sin^-1 (3x – 4x³)
Solution:
Put x = sin θ
y = sin^-1 (3 sin θ – 4 sin³ θ)
= sin^-1 (sin 3 θ) = 3 θ = 3 sin^-1 x.
\(\frac{dy}{dx}\) = \(\frac{3}{\sqrt{1-x^{2}}}\)

ii) cos^-1 (4X3 – 3x)
Solution:
Put x = cos θ
y = cos^-1 (4 cos³ θ – 3 cos θ)
= cos^-1 (cos 3θ) = 3θ = 3 cos^-1 x
\(\frac{dy}{dx}\) = \(\frac{3}{\sqrt{1-x^{2}}}\)

iii) sin^-1 \(\frac{3}{{1-x^{2}}}\)
Solution:
Put x tan θ ⇒ y

iv) tan^-1 \(\frac{a-x}{1+ax}\)
Solution:
Put a tan α, x = tan θ
y = tan^-1 \(\left(\frac{\tan \alpha-\tan \theta}{1+\tan \alpha \tan \theta}\right)\)
= tan^-1 (tan (α – θ)) = α – θ
= tan^-1 a – tan^-1 x;
\(\frac{dy}{dx}\) = 0 – \(\frac{1}{1+x^{2}}\) = – \(\frac{1}{1+x^{2}}\)

v) tan^-1 \(\sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

Differentiating w.r.to x; \(\frac{dy}{dx}\) = \(\frac{1}{2}\)

vi) sin [cos (x²)]
Solution:
\(\frac{dy}{dx}\) = cos [cos (x²)]\(\frac{d}{dx}\) [cos (x²)]
= cos [cos (x²)]. [-sin (x²)] \(\frac{d}{dx}\) (x²)
= cos [cos (x²)] [- sin (x²)]. 2x
= -2x . sin (x²).cos [cos (x²)]

vii) sec^-1 (\(\frac{1}{2x^{2}-1}\)) (0 < x < \(\frac{1}{\sqrt{2}}\))
Solution:
x = cos θ
2x² – 1 = 2 cos² θ – 1 = cos 2θ

viii) sin^-1 [tan^-1 (e^-x)]
Solution:
\(\frac{dy}{dx}\) = cos [tan^-1 (e^-x)]. [tan^-1 (e^-x)]¹
= cos (tan^-1 (e^-x)]x – \(\frac{1}{1+\left(e^{-x}\right)^{2}}\) (e^-x)¹
= \(\frac{-e^{-x}}{1+e^{-2 x}}\) . cos [tan^-1 (e^-x)]

Question 2.
Differentiate f(x) with respect to g(x) for the following.
i) f(x) = e^x, g(x) = √x
Solution:
Let y = e^x and z = √x

ii) f(x) = e^{sin x}, g(x) = sin x.
Solution:
Let y = e^{sin x} and z = sin x.

iii) f(x) = tan^-1 \(\frac{2x}{1-x^{2}}\), g(x) sin^-1 = \(\frac{2x}{1+x^{2}}\)
Solution:
Lety = tan^-1 \(\frac{2x}{1-x^{2}}\), and z = sin^-1 = \(\frac{2x}{1+x^{2}}\)
Put x = tan θ

Question 3.
If y = e^a sin^-1x the prove that \(\frac{dy}{dx}\) = \(\frac{a y}{\sqrt{1-x^{2}}}\)
Solution:
y = e^{a sin-1x}
\(\frac{dy}{dx}\) = e^{a sin-1x} (a sin^-1 x)¹
= e^{a sin-1x} . a \(\frac{1}{\sqrt{1-x^{2}}}\) = \(\frac{a y}{\sqrt{1-x^{2}}}\)

II.

Question 1.
Find the derivatives of the following function.
i) tan^-1 \(\left(\frac{3 a^{2} x-x^{3}}{a\left(a^{2}-3 x^{2}\right)}\right)\)
Solution:
Put x = a tan θ

ii) tan^-1 (sec x + tan x)
Solution:

iii) tan^-1 \(\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Solution:
Put x = tan θ

iv) (logx)^{tan x}
Solution:
log y = log (log x)^{tan x}
= (tan x). log (log x)
Differentiating w.r.to x
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = tan x . \(\frac{d}{dx}\) (log(log x)) + log(logx) \(\frac{d}{dx}\) (tan x)
= tan x. \(\frac{1}{log x}\) . \(\frac{1}{x}\) + log(log x). sec² x

v) (x^x)^x = x^x²
Solution:
log y = log x^x² = x². log x
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = x² . \(\frac{d(\log x)}{d x}\) + (log x ) \(\frac{d}{dx}\) (x²)
= x². \(\frac{1}{x}\) + 2x. log x
= x + 2x log x = x (1 + 2 log x).
= x (log e + log x²)
= x. log (ex²)
\(\frac{dy}{dx}\) = y. x. log (ex²)
= x^x² . x. log (ex²)
= x^{x² +1} log (ex²)

vi) 20^{log (tan x)}
Solution:
log y = log (20)^{log (tan x)}
= log (tan x) log 20
Differentiating w. r. to x

\(\frac{dy}{dx}\) = y. (2 log 20). cosec 2x
= 20^{log (tan x)} (2 log 20). cosec 2x

vii) x^x + e^ex
Solution:
Let y₁ = x^x and y₂ = e^ex so that y = y₁ + y₂.
y₁ = x^x ⇒ log y₁ = log x^x = x log x

viii) x. log x. log (log x)
Solution:
\(\frac{dy}{dx}\) = x. log x \(\frac{d}{dx}\) (log. (log x)) + log (log x) logx. 1 + x. log (log x) \(\frac{1}{x}\).
= x log x. \(\frac{1}{log x}\) . \(\frac{1}{x}\) + log x. log (log x) + log (log x)
= 1 + log (logx) (1 + logx) = 1 + log (logx) + log x log (log x)
= log e + log (log x) + log x. log (log x)
= log (e log x) + log x. log (log x)

ix) e^-ax² sin (x log x)
Solution:
\(\frac{dy}{dx}\) = e^-ax² . \(\frac{d}{dx}\) (sin (x log x)) dx + sin (x log x) \(\frac{d}{dx}\) (e^-ax²)
= e^-ax² cos (x log x). (x . \(\frac{1}{x}\) + log x) + sin (x log x) e^-ax² (-2ax)
= e^-ax² [(cos (x log x) (1 + log x) – 2 ax. sin (x log x)]
= e^-ax² (cos (x log x) (log ex) -2 ax. sin (x log x))

x) sin^-1\(\left(\frac{2^{x+1}}{1+4^{x}}\right)\) (Put 2ⁿ = tan θ)
Solution:
sin^-1\(\left(\frac{2^{x+1}}{1+4^{x}}\right)\)
Put 2^x = tan θ.

Question 2.
Find \(\frac{dy}{dx}\) for the following functions.
i) x = 3 cos t – 2 cos³ t,
y = 3 sin t – 2 sin³ t
Solution:
\(\frac{dx}{dt}\) = – 3 sin t – 2(3 cos² t) (- sin t)
= – 3 sin t + 6 cos² t (sin t)
= 3 sin t (2 cos² t – 1)
= 3 sin t. cos 2t
y = 3 sin t – 2 sin³ t
\(\frac{dy}{dt}\) = 3 cos t – 2 (3 sin² t) (- cos t)
= 3 cos t (1 – 2 sin² t)
= 3 cos t. cos 2t
\(\frac{dy}{dx}\) = \(\frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}=\frac{3 \cos t \cos 2 t}{3 \sin t \cos 2 t}\) = cot t

ii) x = \(\frac{3 a t}{1+t^{3}}\), y = \(\frac{3 a t^{2}}{1+t^{3}}\)
Solution:

iii) x = a (cos t + t sin t), y = a (sin t – t cos t)
Solution:
\(\frac{dx}{dt}\) = a (- sin t + t cos t + sin t) = at cos t
∴ y = a (sin t – t cos t)
\(\frac{dy}{dt}\) = a (cos t – cos t + t sin t) = at sin t
\(\frac{dy}{dx}\) = \(\frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}=\frac{at\cos t}{at\cos t}\) = tan t

iv) x = a\(\left[\frac{\left(1-t^{2}\right)}{1+t^{2}}\right], \mathbf{y}=\frac{2 b t}{1+t^{2}}\)
Solution:

Question 3.
Differentiate f(x) with respect to g(x) for the following.
i) f(x) = log_a^x, g(x) = a^x.
Solution:
y = log_a^x = \(\frac{\log x}{\log _{c}^{a}}\)

ii) f(x) = sec^-1 (\(\frac{1}{2x^{2}-1}\)) g(x) = \(\sqrt{1-x^{2}}\)
Solution:
Let y = sec^-1 (\(\frac{1}{2x^{2}-1}\)) and z = \(\sqrt{1-x^{2}}\)
Put x = cos θ

iii) f(x) = tan^-1\(\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\), g(x) = tan^-1 x.
Solution:
Let y = tan^-1\(\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and z = tan^-1 x
x = tan z

Question 4.
Find the derivative of the function y defined implicitly by each of the following equations.
i) x⁴ + y⁴ – a² xy = 0
Solution:
Differentiate w.r.to x
4x³ + 4y³ . \(\frac{dy}{dx}\) – a²(x. \(\frac{dy}{dx}\) + y . 1 = 0)
4x³ + 4y³ . \(\frac{dy}{dx}\) – a²x\(\frac{dy}{dx}\) – a²y = 0
4y³ – a²x) \(\frac{dy}{dx}\) = a²y – 4x³ ; \(\frac{dy}{dx}\) = \(\frac{a^{2} y-4 x^{3}}{4 y^{3}-a^{2} x}\)

ii) y = x^y
Solution:
log y = log x^y = y log x
Differentiate w.r.to x

iii) y^x = x^{sin y}
Solution:
Take log on both sides
log y^x = log x^{sin y} ⇒ x. log y = (sin y) log x.
Differentiating w.r.to x

Question 5.
Establish the following.
i) If \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\) = a(x – y), than \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}\)
Solution:
Given \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\) = a(x – y)
Put x = sin θ, y = sin Φ
Differentiating w.r. to x

ii) If y = x \(\sqrt{a^{2}+x^{2}}\) + a² log (x + \(\sqrt{a^{2}+x^{2}}\)), then \(\frac{dy}{dx}\) = 2 \(\sqrt{a^{2}+x^{2}}\)
Solution:
y ⇒ x\(\sqrt{a^{2}+x^{2}}\) + a² log (x + \(\sqrt{a^{2}+x^{2}}\))

iii) If x^{log y} = log x, then
Solution:
Given x^{log y} = log x, log x^{log y} = log log x
(log y) (log x) = log(logx).
Differentiating w.r.to x

iv) If y = Tan^-1 \(\frac{2x}{1-x^{2}}\) + Tan^-1 \(\frac{3x-x^{3}}{1-3x^{2}}\) – tan^-1 \(\frac{4x-4x^{3}}{1-6x^{2}+x^{4}}\) than \(\frac{dy}{dx}\) = \(\frac{1}{1+x^{2}}\)
Solution:
Put x = tan θ

= tan^-1 (tan 2θ) + tan^-1 (tan 3θ) – tan^-1 (tan 4θ)
= 2θ + 3θ – 4θ = θ = tan^-1 x
∴ \(\frac{dy}{dx}\) = \(\frac{1}{1+x^{2}}\)

v) If x^y = y^x, then \(\frac{dy}{dx}\) = \(\frac{y(x log y – y)}{x(y log x – x)}\)
Solution:
Given x^y = y^x ; log x^y = log y^x
y log x = x log y
Differentiating w.r.to x

vi) If x^2/3 + y^2/3 = a^2/3 then \(\frac{dy}{dx}\) = -3 √y/x
Solution:
Given x^2/3 + y^2/3 = a^2/3
Differentiating w.r.to x

Question 6.
Find the derivative \(\frac{dy}{dx}\) of each of the following functions.
i) y = \(\frac{(1-2 x)^{2 / 3}(1+3 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}}\)
Solution:
log y = log \(\left\{\frac{(1-2 x)^{2 / 3}(1+3 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}}\right\}\)
= log (1 – 2x)^2/3 + log (1 + 3x)^-3/4 – log (1 – 6x)^5/6 – log (1 + 7x)^-6/7
= \(\frac{2}{3}\) log (1 – 2x) – \(\frac{3}{4}\) log (1 + 3x) – \(\frac{5}{6}\) log (1 – 6x) + \(\frac{6}{7}\) log (1 + 7x)
Differentiating w.r.to x
\(\frac{1}{y}\).\(\frac{dy}{dx}\) = \(\frac{2}{3}\) . \(\frac{1(-2)}{1-2x}\) – \(\frac{3}{4}\) . \(\frac{1}{1+3x}\) . 3

ii)

Solution:
log y = log x⁴ + log (x² + 4)^1/3 – log (4x² – 7)^1/2
= 4 log x + \(\frac{1}{3}\) log (x² + 4) – \(\frac{1}{2}\) log (4x² – 7)
Differentiating w.r.to x

iii) y = \(\frac{(a-x)^{2}(b-x)^{3}}{(c-2 x)^{3}}\)
Solution:
log y = log \(\frac{(a-x)^{2}(b-x)^{3}}{(c-2 x)^{3}}\)
= log (a – x)² + log (b – x)³ – log (c – 2x)³
= 2 log (a – x) + 3 log (b – x) – 3 log (c – 2x)
Differentiating w.r.to x

iv)

Solution:
log y = log \(\frac{x^{3}(2+3 x)^{1 / 2}}{(2+x)(1-x)}\)
= log x³ + log (2 + 3x)^1/2 – log (2 + x) – log (1 – x)
= 3 log x + \(\frac{1}{2}\) log (2 + 3x) – log (2 + x) – log (1 – x)
Differentiating w.r. to x

v) y = \(\sqrt{\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}}\)
Solution:
log y = log(\(\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}\))^1/2
= \(\frac{1}{2}\) log \(\frac{(x-3)\left(x^{2}+4\right)}{3 x^{2}+4 x+5}\)
= \(\frac{1}{2}\) (log (x – 3) + log (x² + 4) – log (3x² + 4x + 5))

III.

Question 1.
Find the derivatives of the following functions.
i) y = (sin x)^logx + x^{sin x}
Solution:
Let y₁ =(sinx)^logx, y₂ = x^{sin x} so that y = y₁ + y₂
y₁ = (sin x)^logx
log y₁= log{ (Sin x)^logx} = log x. log (sin x)
Differentiating w.r. to x

y₂ = x^{sin x}
log y₂ = (log x)^{sin x} = sin x. logx

ii) x^xx
Solution:
log y = log x(x^xx) = x^x. log X
\(\frac{1}{y}\) \(\frac{dy}{dx}\) = x^x. \(\frac{1}{x}\) + (log x). x^x (1 + log x)
[\(\frac{d}{dx}\)(x^x) = x^x (1 + log x)]
= x^x-1 [1+ x log x (log e + log x)]
= x^x-1 (1 + x. log x. log ex)
\(\frac{dy}{dx}\) = y.x^x-1 (1 + x log x. log ex)
= x^(xx) . x^x-1 (1 + x log x. log ex)
= x^xx+x-1 (1 + x log x. log ex)

iii) (sin x)^x + x^{sin x}
Solution:
Let y₁ = (sin x)^x and y₂ = x^{sin x}
so that y = y₁ + y₂
log y₁ = log (sin x)^x = x. log sin x

iv) x^x + (cot x)^x
Solution:
Let y₁ = x^x and y₂ = (cot x)^x
log y₁ = log x^x = x log x

Question 2.
Establish the following
i) If x^y + y^x = a^b then
\(\frac{dy}{dx}\) = \(-\left(\frac{y \cdot x^{y-1}+y^{x} \cdot \log y}{x^{y} \cdot \log x+x \cdot y^{x-1}}\right)\)
Solution:
Let y₁ = x^y and y₂ = y^x. so that y₁ + y₂ = a^b
logy₁ = log x^y = y logx

ii) If f(x) = sin^-1\(\sqrt{\frac{x-\beta}{\alpha-\beta}}\) and
g(x) = tan \(\sqrt{\frac{x-\beta}{\alpha-x}}\) than
f'(x) = g'(x) (β < x < α)
Solution:

iii) If a > b > 0 and 0 < x < π
f(x) = (a – b)^-1/2 . cos^-1\(\left(\frac{a \cos x+b}{a+b \cos x}\right)\), than f'(x) = (a + b cos x)^-1
Solution:
Let u = cos^-1\(\left(\frac{a \cos x+b}{a+b \cos x}\right)\)

Question 3.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) by
i) Using product
ii) Obtaining a single polynomial expanding the product
iii) Logarithmic differentiation do they all give the same answer?
Solution:
Do Product rule:
y = (x² – 5x + 8) (x³ + 7x + 9)
\(\frac{dy}{dx}\) = (x² – 5x + 8) \(\frac{d}{dx}\)(x³ + 7x + 9) + (x³ + 7x + 9) \(\frac{d}{dx}\)(x² – 5x + 8)
= (x² – 5x + 8)(3x² + 7) + (x³ + 7x + 9)(2x – 5)
= 3x⁴ – 15x³ + 24x² + 7x² – 35x + 56 + 2x⁴ + 14x² + 18x – 15x³ – 35x – 45
= 5x⁴ – 20x³ + 45x² – 52x + 11 ……….. (1)

ii) Expanding the product :
Solution:
y = (x² – 5x + 8) (x³ + 7x + 9)
= 5x⁵ + 7x³ + 9x² – 5x⁴ -35x² – 45x + 8x³ + 56x + 72
= x⁵ – 5x⁴ + 15x³ – 26x² + 11x +72
\(\frac{dy}{dx}\) = 5x⁴ – 20x³ + 45x² – 52x + 11 ……….. (2)

iii) y = (x² – 5x + 8) (x³ + 7x + 9)
Solution:
log y = log (x² – 5x + 8) (x³ + 7x + 9)
= log (x² – 5x + 8) + log (x³ + 7x + 9)

= (2x – 5)(x³ + 7x + 9) + (x² – 5x + 8)(3x² + 7)
= 2x⁴ + 14x² + 18x – 5x³ – 35x – 45 + 3x⁴ -15x³ + 24x² + 7x² – 35x + 56
= 5x⁴ – 20x³ +45x² – 52x + 11 ……….. (3)
From (1), (2) and (3) we observe that all the three give same answer.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(b)

I. Compute the following limits.

Question 1.
Find the derivatives of the following function.
i) cotⁿ x
Solution:
\(\frac{dy}{dx}\) = n. cot^n-1 x. \(\frac{d}{dx}\) (cot x)
= n. cot^n-1 x (- cosec² x)
= – n. cot^n-1 x. cosec² x

ii) cosec⁴ x
Solution:
\(\frac{dy}{dx}\) = 4.cosec³ x. \(\frac{d}{dx}\)(cosec x)
= 4. cosec³ x (- cosec x. cot x)
= – 4. cosec⁴ x. cot x

iii) tan (e^x)
=sec 2(e^x).(e^x)¹
= e^x. sec² (e^x)

iv) \(\frac{1-\cos 2 x}{1+\cos 2 x}\)
Solution:
\(\frac{2 \sin ^{2} x}{2 \cos ^{2} x}\) = tan² x \(\frac{dy}{dx}\) = 2 tan x . sec² x

v) sin^m x cosⁿ x
Solution:
\(\frac{dy}{dx}\) = (sin^m x). \(\frac{d}{dx}\) (cosⁿ x) + (cosⁿ x) \(\frac{d}{dx}\) (sin^m x)
= sin^m xn + cos^n-1 x(-sin x) + cosⁿ x. m sin^m-1 x. cos x
= m. cosⁿ⁺¹ x. sin^m-1 x – n. sin^m+1 x. cos^n-1 x.

vi) sin mx. cos nx
Solution:
\(\frac{dy}{dx}\) = sin mx \(\frac{d}{dx}\) (cos nx) + (cos nx) \(\frac{dy}{dx}\) (sin mx)
= sin mx (-n sin nx)+cos nx (m cos mx)
= m. cos mx. cos nx – n. sin mx . sin nx

vii) x tan^-1 x
Solution:
\(\frac{dy}{dx}\) = x. \(\frac{d}{dx}\) (tan^-1 x) + (tan^-1 x) \(\frac{d}{dx}\) (x)
= \(\frac{x}{1+x^{2}}\) + tan^-1 x.

viii) sin^-1 (cos x)
Solution:
= sin^-1 [sin (\(\frac{\pi}{2}\) – x)] = \(\frac{\pi}{2}\) – x
\(\frac{dy}{dx}\) = 0 – 1 = -1

ix) log (tan 5x)
Solution:

x) sinh^-1 (\(\frac{3x}{4}\))
Solution:

xi) tan^-1 (log x)
Solution:

xii) log (\(\frac{x^{2}+x+2}{x^{2}-x+2}\))
Solution:
\(\frac{dy}{dx}\) = log(x² + x + 2) – log(x² – x + 2)

xiii) log (sin^-1 (e^x))
Solution:

xiv) (sin x)² (sin^-1 x)²
Solution:

xv) y = \(\frac{\cos x}{\sin x+\cos x}\)
Solution:

xvi) \(\frac{x\left(1+x^{2}\right)}{\sqrt{1-x^{2}}}\)
Solution:

xvii) y = e^sin-1x
Solution:

xviii) y = cos (log x + e^x)
Solution:
\(\frac{dy}{dx}\) = -sin(log x + e^x) = \(\frac{d}{dx}\) (log x + e^x)
= -sin (log x + e^x) (\(\frac{1}{x}\) + e^x)

xix) y = \(\frac{\sin (x+a)}{\cos x}\)
Solution:

xx) y = cot^-1 (coses 3x)
Solution:

Question 2.
Find the derivatives of the following fountion.
i) x = sinh² y
Solution:
\(\frac{dy}{dx}\) = 2 sinh y . cosh y

ii) x = tanh² y
Solution:
\(\frac{dy}{dx}\) = 2 tanh y . sech² y

iii) x = e^{sinh y}
Solution:
\(\frac{dy}{dx}\) = e^{sinh y} \(\frac{d}{dx}\) (sinh y)
= e^{sinh y} . cosh y
= x. cosh y
\(\frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{x \cdot \cosh y}\)

iv) x =tan (e^-y)
Solution:
\(\frac{dy}{dx}\) = sec² (e^-y) . (e^-y)¹ = -e^-y . sec² e^-y
= -e^-y(1 + tan² (e^-y) = – e^-y(1 + x²)
\(\frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=-\frac{1}{e^{-y}\left(1+x^{2}\right)}=-\frac{e^{y}}{1+x^{2}}\)

v) x = log (1 + sin² y)
Solution:

vi) x = log (1 + √y)
Solution:
1 + √y = e^x
√y = e^x – 1
y = (e^x – 1)²
\(\frac{dy}{dx}\) = 2(e^x – 1) . e^x = 2 √y . e^x
= 2 √y (√y + 1)
= 2(y + √y)

II. Find the derivativies of the following functions.

i) y = cos [log (cot x)]
Solution:

ii) y = sin^-1 \(\frac{1-x}{1+x}\)
Solution:

iii) log (cot (1 – x²))
Solution:

iv) y = sin [cos (x²)]
Solution:
\(\frac{dy}{dx}\) = cos [cos (x²)].\(\frac{d}{dx}\)[cos (x²)]
= cos [cos (x²)](sin (x²)).\(\frac{d}{dx}\)(x²)
= – 2x. sin (x²). cos [cos (x²)]

v) y = sin [tan^-1 (e^x)]
Solution:

vi) y = \(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Solution:

vii) y = tan^-1 (tan h \(\frac{x}{2}\))
Solution:

viii) y = sinx . (Tan^-1x)²
Solution:

III. Find the derivatives of the following functions.

Question 1.
y = sin^-1 \(\left(\frac{b+a \sin x}{a+b \sin x}\right)\) (a > 0, b > 0)
Solution:
Let u = \(\frac{b+a \sin x}{a+b \sin x}\)

Question 2.
cos^-1\(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) (a > 0, b > 0)
Solution:
Let u = \(\frac{b+a \cos x}{a+b \cos x}\)

Question 3.
tan^-1 \(\left[\frac{\cos x}{1+\cos x}\right]\)
Solution:
Let u = \(\frac{\cos x}{1+\cos x}\)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(a)

I. Compute the following limits.

Question 1.
Find the derivatives of the following functions f(x).
i) \(\sqrt{x}+2 x^{\frac{3}{4}}+3 x^{\frac{5}{6}}\) (x > 0)
Solution:
y = \(\sqrt{x}+2 x^{\frac{3}{4}}+3 x^{\frac{5}{6}}\) (x>0)
\(\frac{dy}{dx}\) = \(\frac{1}{2}\).x^-1/2 + 2.\(\frac{3}{4}\).x^-1/4 + 3.\(\frac{5}{6}\).^x-1/6 dx
= \(\frac{1}{2}\)[x^-1/2 + 3.x^-1/4 + 5.x^-1/6]

ii) \(\sqrt{2 x-3}+\sqrt{7-3 x}\).
Solution:

iii) (x² – 3) (4x³ + 1)
Solution:
y = (x² – 3) (4x³ + 1)
\(\frac{dy}{dx}\) = (x² – 3) \(\frac{d}{dx}\) (4x³ + 1) + (4x³ + 1) \(\frac{d}{dx}\)(x² – 3)
= (x² – 3) (12x²) + (4x³ + 1) (2x)
= 12x⁴ – 36x² + 8x⁴ + 2x
= 20x⁴ – 36x² + 2x

iv) (√x – 3x) (x + \(\frac{1}{x}\))
Solution:
y = (√x – 3x) (x + \(\frac{1}{x}\))

v) (√x + 1) (x² – 4x + 2) (x > 0)
Solution:
y = (√x + 1) (x² – 4x + 2) (x > 0)
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (√x + 1) \(\frac{d}{dx}\)(x² – 4x + 2) + (x² – 4x + 2) \(\frac{d}{dx}\)(√x +1)
= (√x + 1) (2x – 4) + \(\frac{x^{2}-4 x+2}{2 \sqrt{x}}\)

vi) (ax + b)ⁿ. (cx + d)^m.
Solution:
y = (ax + b)ⁿ. (cx + d)^m
\(\frac{dy}{dx}\) = (ax + b)ⁿ \(\frac{d}{dx}\)(cx +d)^m + (cx + d)^m \(\frac{d}{dx}\)(ax + b)ⁿ
= (ax + b)ⁿ [m(cx + d)^m-1. c] + (cx + d)^m [n(ax + b)^n-1. a]
= (ax + b)^n-1 (cx + d)^m-1 [cm (ax + b) + an (cx + d)]
= (ax + b)ⁿ (cx + d)^m [\(\frac{an}{ax+b} + \frac{cm}{cx+d}\) ]

vii) 5 sin x + ex log x
Solution:
y = 5 sin x + e^x. log x
\(\frac{dy}{dx}\) = 5 cos x + e^x. \(\frac{d}{dx}\) (log x) + log x \(\frac{d}{dx}\)(e^x)
= 5 cos x + e^x. \(\frac{1}{x}\) + (log x) (e^x)

viii) 5^x + log x + x³ e^x
Solution:
y = 5^x + log x + x³ e^x
\(\frac{dy}{dx}\) = 5^x . log 5 + \(\frac{1}{x}\) + x³.e^x + e^x.3x²
= 5^x. log 5 + – + x³ e^x + 3x² e^x

ix) e^x + sin x cos x
Solution:
y = e^x + sin x . cos x
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) (e^x) + \(\frac{d}{dx}\) (sin x . cos x)
= e^x + sin x \(\frac{d}{dx}\) (cos x) + cos x \(\frac{d}{dx}\) (sin x)
= e^x – sin² x + cos² x
= e^x + cos 2x

x) \(\frac{p x^{2}+e x+r}{a x+b}\)(|a| + |b| ≠ 0)
Solution:

xi) log₇ (log x) (x > 0)
Solution:
y = log₇ (log x) (x > 0)

xii) \(\frac{1}{a x^{2}+b x+c}\) (|a| + |b| + |c| ≠ 0)
Solution:
\(\frac{1}{a x^{2}+b x+c}\) (|a| + |b| + |c| ≠ 0)

xiii) e^2x log (3x + 4) (x > –\(\frac{4}{3}\))
Solution:
y = e^2x. log (3x + 4) (x > –\(\frac{4}{3}\))
Differentiating w.r.to x
\(\frac{dy}{dx}\) = e^2x \(\frac{d}{dx}\)[log (3x + 4) + log (3x + 4) \(\frac{d}{dx}\) (e^2x)]
= e^2x.\(\frac{1}{3x+4}\) 3 + log (3x + 4). e^2x . 2
= e^2x (\(\frac{3}{3x+4}\) + 2 log (3x + 4))

xiv) (4 + x²) e²xy
Solution:
y = (4 + x²). e^2x
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (4 + x²) \(\frac{d}{dx}\) (e^2x) + e^2x \(\frac{d}{dx}\)(4 + x²)
= (4 + x²). 2e^2x + e^2x (0 + 2x)
= 2e^2x [4 + x² + x]
= 2e^2x (x² + x + 4)

xv) \(\frac{ax+b}{cx+d}\) [|c| + |d|≠0]
Solution:
y = \(\frac{ax+b}{cx+d}\) [|c| + |d|≠0]
Differentiating w.r.to x

xvi) a^x. e^x²
Solution:
y = a^x. e^x²
Differentiating w.r.to x
\(\frac{dy}{dx}\) = (a^x) \(\frac{d}{dx}\)(e^x²) + (e^x²)\(\frac{d}{dx}\)(a^x)
= a^x. e^x². 2x + e^x². a^x. log a
= ax. ^x² (2x + log a)
= y(2x + log a)

Question 2.
If f(x) = 1 + x + x² + + x¹⁰⁰, then find f’ (1).
Solution:
f'(x) = 1 + 2x + 3x² + 100 x⁹⁹
f'(1) = 1+2 + 3 ….+ 100
= \(\frac{100 \times 101}{2}=5050\left(\Sigma x=\frac{x(x+1)}{2}\right)\)

Question 3.
If f (x) = 2x² + 3x – 5, then prove that f(0) + 3f (-1) = 0.
Solution:
f'(x) = 4x + 3
f'(0) = 0 + 3 = 3
f'(-1) = – 4 + 3 = -1
f'(0) + 3f'(-1) = 3 + 3(-1) = 3 – 3 = 0 n.

II.

Question 1.
Find the derivatives of the following functions from the first principles.
i) x³
Solution:


= 3x² + 0 + 0 = 3x²

ii) x⁴ + 4
Solution:
f(x + h) – f(x) = ((x + h)⁴ + 4) – (x⁴ + 4)
= (x + h)4 + 4 – x⁴ – 4 .
= x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴ – x⁴

= 4x³ + 0 + 0 + 0 = 4x³

iii) ax² + bx + c
Solution:
f(x + h) = a(x + h)² + b(x + h) + c
= a(x² + 2hx + h²) + b(x + h) + c
= ax² + 2ahx + ah² + bx + bh +c

f(x + h) – f(x) = ax² + 2ahx + ah² + bx + bh + c – ax² – bx – c
= h [2ax + ah + b]

= 2ax + 0 + b = 2ax + b

iv) \(\sqrt{x+1}\)
Solution:

v) sin 2x
Solution:
f(x + h) – f(x) = sin 2(x + h) – sin 2x

vi) cos ax
Solution:
f(x + h) – f(x) = cos a (x + h) – cos ax

= – 2 sin ax. \(\frac{a}{2}\)
=-a. sin ax

vii) tan 2x
Solution:
f(x + h) – f(x) = tan 2(x + h) – tan 2x

viii) cot x
Solution:
f(x + h) – f(x) = cot (x + h) – cot x

ix) sec 3x
Solution:
f(x + h) – f(x)= sec 3(x + h) – sec 3x

x) x sin x
Solution:
f(x + h) – f(x) = (x + h) sin (x + h) – x sin x
= x (sin (x + h) – sin x) + h. sin (x + h)

xi) cos² x
Solution:
f(x + h) – f(x) = cos² (x + h) – cos² x
= -(cos² x – cos² (x + h))
= -sin (x + h + x) sin (x + h – x)

= -sin 2x. 1 = -sin 2x

Question 2.
Find the derivatives of the following function.
i) \(\frac{1-x \sqrt{x}}{1+x \sqrt{x}}\) (x > 0)
Solution:

ii) xⁿ n^x log (nx) (x > 0, n ∈ N)
Solution:
y = xⁿ. n^x. log nx
\(\frac{dy}{dx}\) = xⁿ. n^x (log nx) + nⁿ. log xn (xⁿ) + xⁿ. log nx (n^x)
= xⁿ. n^x \(\frac{n}{log nx}\) + n^x. log xⁿ (nx^n-1) + xⁿ . log nx. (n^x . log nx)
= x^n-1. n^x[\(\frac{nx}{log nx}\) + log nx. (nⁿ. log nx)]

iii) ax²ⁿ. log x + bxⁿ e^-x
Solution:
y = ax²ⁿ. log x + bxⁿ e^-x
\(\frac{dy}{dx}\) = a (x²ⁿ.\(\frac{1}{x}\) + log x (2nx^2n-1)) + b n (-e^-x) + e^-x. nx^n-1)
= a. x^2n-1 + 2an. x^2n-1. log x – bxⁿ e^-x + bn. x^n-1 . e^-x

iv) (\(\frac{1}{x}\) – x)³ e^x
Solution:
y = (\(\frac{1}{x}\) – x)³ . e^x
\(\frac{dy}{dx}\) = (\(\frac{1}{x}\) – x)³ \(\frac{d}{dx}\)(e^x) + e^x \(\frac{d}{dx}\){(\(\frac{1}{x}\) – x)³}

Question 3.
Show that the function f(x) = |x| + |x – 1|, x ∈ R is differentiable for all real numbers except for 0 and 1.
Solution:
f(x) = |x| + |x – 1| ∀ x ∈ R
f(x) = x + x – 1 = 2x – 1, x ≥ 1
= x – (x – 1) = x – x + 1, = 1, 0 < x < 1
= -x – (x – 1) =-x – x + 1 = 1 – 2x, x < 0 ∴ f(x) = 2x – 1, x > 1
= 1, 0 < x < 1 = 1 – 2x, x ≤ 0 If x > 1, then f(x) = 2x – 1 = polynomial in x f(x) is differentiable for all x > 1
If 0 < x < 1, then f(x) = 1 – constant
∴ f(x) is differentiable if 0 < x < 1.
If x < 1, then f(x) = 1 – 2x = polynomial in x.
∴ f(x) is differentiable for all x < 1

Case (i) : x = 0

R f'(0) ≠ Lf'(0)
∴ f'(0) does not exist.
f(x) is not differentiable at x = 0.

Case (ii): x = 1

R f'(1) ≠ L f'(1)
f(1) does not exist.
f(x) is not differentiable at x = 1
∴ f(x) is differentiable for all real x except zero and one.

Question 4.
Verify whether the following function is differentiable at 1 and 3.

Solution:
Case (i): x = 1

R f'(1) ≠ L f'(1)
f(x) is not differentiable at x = 1

Case (ii) : x = 3

f(3⁺) ≠ f'(3^–)
f(x) is not differentiable at x = 3.

Question 5.
Is the following function f derivable at 2?

Solution:

f'(2^–) ≠ f(2⁺); f(x) is not derivable at x = 2.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(e)

I.

Question 1.
Is the function f, defined by \(f(x)=\left\{\begin{array}{l}
x^{2} \text { if } x \leq 1 \\
x \text { if } x>1
\end{array}\right.\) continuous on R?
Solution:

f is continuous at x = 1
f is continuous on R.

Question 2.
Is f defined by f(x) = \(=\left\{\begin{array}{cc}
\frac{\sin 2 x}{x}, & \text { if } x \neq 0 \\
1 & \text { if } x=0
\end{array}\right.\) continuous at 0?
Solution:

f is not continuous at 0

Question 3.
Show that the function f(x) = [cos (x¹⁰ + 1)]^1/3, x ∈ R is a continuous function.
Solution:
We know that cos x is continuous for every x ∈ R
∴ The given function f(x) is continuous for every x ∈ R.

II.

Question 1.
Check the continuity of the following function at 2.

Solution:

f(x) is not continuous at 2.

Question 2.
Check the continuity of f given by f(x) = \(\begin{cases}\frac{\left[x^{2}-9\right]}{\left[x^{2}-2 x-3\right]} & \text { if } 0<x<5 \text { and } x \neq 3 \\ 1.5 & \text { if } x=3\end{cases}\) at the point 3.
Solution:

f(x) is continuous at x = 3.

Question 3.
Show that f, given by f(x) = \(\frac{x-|x|}{x}\) (x ≠ 0) is continuous on R – {0}.
Solution:
Case (i) : a > 0 ⇒ |a| = a

If x = 0, f(a) is not defined
f(x) is not continuous at ’0′
∴ f(x) is continuous on R – {0}

Question 4.
If f is a function defined by

then discuss the continuity of f.
Solution:
Case (i) : x = 1

f(x) is not continuous at x > 1

Case (ii) : x = -2

f(x) is not continuous at x = -2.

Question 5.
If f is given by f(x) = \(=\left\{\begin{array}{cl}
k^{2} x-k & \text { if } x \geq 1 \\
2 & \text { if } x<1
\end{array}\right.\) is a continuous function on R, then find the values of k.
Solution:

2 = k² – k
k² – k – 2 = 0
(k – 2) (k + 1) = 0
k = 2 or – 1

Question 6.
Prove that the functions ‘sin x’ and ‘cos x’ are continuous on R.
Solution:
i) Let a ∈ R

∴ f is continuous at a.

ii) Let a ∈ R

∴ f is continuous at a.

III.

Question 1.
Check the continuity of ‘f given by

at the points 0, 1 and 2.
Solution:
i)
∴ f(x) is continuous at x = 0

ii)
∴ f(x) is continuous at x = 1

iii)
∴ f(x) is continuous at x = 2

Question 2.
Find real constant a, b so that the function f given by

is continuous on R.
Solution:

Since f(x) is continuous on R
LHS = RHS ⇒ a = 0

Since f(x) is continuous on R.
LHS = RHS
3b + 3 = -3
3b = – 6 ⇒ b = -2

Question 3.
Show that

where a and b are real constants, is continuous at 0.
Solution:

∴ f(x) is continuous at x = 0

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(d)

I. Compute the following limits.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

II.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

III.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(c)

I. Computer the following limits.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:
As x → 0
⇒ 7x → 0

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

II.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:
By L. Hospital rule

III.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Limits and Continuity Solutions Exercise 8(b)

Find the right and left-hand limits of the functions in 1,2,3 of I and 1,2, of II at the point mentioned against them. Hence, check whether the functions have limits at those a’s.

I.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

II.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:
x → 2 – ⇒ x < 2
x – 2 < 0
|x – 2| = – (x – 2)

Question 4.

Solution:
x → 0 + ⇒ x > 0
|x| = x

Question 5.

Solution:

Question 6.

Solution:

III.

Question 1.

Solution:

∴ LHL ≠ RHL
Limit does not exists.

Question 2.

Solution: