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

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

I.

Question 1.
Without using the derivative, show that
i) The function f(x) = 3x + 7 is strictly increasing on R.
Solution:
Let x1, x2 ∈ R with x1 < x2
Then 3x1 < 3x2
3x1 + 7 < 3x2 + 7
⇒ f(x1) < f(x2)
∴ x1 < x2 ⇒ f(x1) < f(x2) ∀ x1 x2 ∈ R
∴ The given function is strictly increasing on R

ii) The function f(x) = ($$\frac{1}{2}$$)x is strictly decreasing on R.
Solution:
f(x) = ($$\frac{1}{2}$$)x
Let x1, x2 ∈ R
Such that x1 < x2
⇒ ($$\frac{1}{2}$$)x1 > ($$\frac{1}{2}$$)x2
⇒ f(x1) > f(x2)
∴ f(x) is strictly decreasing on R.

iii) The function f(x) = e3x is strictly increasing on R.
Solution:
f(x) = e3x
Let x1, x2 ∈ R such that x1 < x2
We know that of a > b then ea > eb
Then e3x < e3x2
⇒ f(x1) < f(x2)
∴ f is strictly increasing on R.

iv) The function f(x) = 5. – 7x is strictly decreasing on R.
Solution:
f(x) = 5 – 7x
Let x1 x2 ∈ R
Such that x1 < x2
Then 7x1 < 7x2
-7x1 > -7x2
5 – 7x1 > 5 – 7x2
f(x1) > (x2)
∴ x1 < x2 ⇒ f(x1) > f(x2) V x1 x2 ∈ R.
The given function f is strictly decreasing on R. Question 2.
Show that the function f(x) = sin x. Define on R is neither increasing nor decreasing on (0, π).
Solution:
f(x) = sin x
Since 0 < x < n
Consider 0 < x
f(0) < f(x)
sin 0 < sin x
0 < sin x ……….. (1)
Consider x < n
f(x) < f(π)
sin x < sin π 0 > sin x ………….. (2)
From (1) & (2); f(x) is neither increasing nor decreasing.

II.

Question 1.
Find the intervals in which the following functions are strictly increasing or strictly decreasing.
i) x² + 2x – 5
Solution:
Let f(x) = x² + 2x – 5
f'(x) = 2x + 2
f(x) is increasing if f'(x) > 0
⇒ 2x + 2 < 0 ⇒ x+ 1 > 0
x > -1
f(x) is increasing If x ∈ (-1, ∞)
f(x) is decreasing If f'(x) < 0
⇒ 2x + 2 < 0
⇒ x + 1 < 0
⇒ x < -1 f(x) is decreasing if x ∈ (-∞, -1).

ii) 6 – 9x – x².
Solution:
Let f(x) = 6 – 9x – x²
f'(x) = -9 – 2x
f(x) is increasing if f'(x) > 0
⇒ -9 -2x > 0
⇒ 2x + 9 < 0
x < $$\frac{-9}{2}$$
f(x) is increasing if x ∈ (-∞, $$\frac{-9}{2}$$)
f(x) is decreasing if f'(x) < 0
⇒ 2x + 9 > 0
⇒ x > $$\frac{-9}{2}$$
f(x) is decreasing of x ∈ ($$\frac{-9}{2}$$, ∞)

iii) (x + 1)³ (x – 1)³.
Solution:
Let f(x) = (x + 1)³ (x – 1)³
= (x² – 1)³
x6 – 1 – 3x4 + 3x ²
f'(x) = 6x5 – 12x³ + 6x
= 6(x5 – 2x³ + x)
= 6x(x4 – 2x² +1)
= 6x(x² – 1)²
f'(x) ≤ 0
⇒ 6x(x² – 1)² < 0
f(x) is decreasing when (-∞, -1) ∪ (-1, 0)
f'(x) > 0
f(x) is increasing when (0, 1) ∪ (1, ∞)

iv) x³(x – 2)²
Solution:
f'(x) = x³. 2(x – 2) + (x – 2)².3x²
= x² (x – 2) [2x + 3 (x- 2)]
= x² (x – 2) (2x + 3x – 6)
= x² (x – 2) (5x – 6) ∀ x ∈ R, x² ≥ 0
For increasing, f'(x) = 0
x²(x – 2) (5x – 6) > 0
x ∈ (-∞, $$\frac{6}{5}$$) ∪ (2, ∞)
For decreasing, f'(x) < 0
x²(x – 2) (5x – 6) < 0
x ∈ ($$\frac{6}{5}$$, 2)

v) xex
Solution:
f'(x) = x . ex + ex. 1 = ex(x + 1)
ex is positive for all real values of x
f'(x)>0 ⇒ x + 1 > 6 ⇒ x > – 1
f(x) is increasing when x > – 1
f(x) < 0 ⇒ x + 1 < 0 ⇒ x < -1
f(x) is decreasing when x < – 1

vi) $$\sqrt{(25-4x^{2})}$$
Solution:
f(x) is real only when 25 – 4x² > 0
-(4x² – 25) > 0
-(2x + 5) (2x – 5) > 0
∴ x lies between –$$\frac{5}{2}$$ and $$\frac{5}{2}$$
Domain of f = (-$$\frac{5}{2}$$, $$\frac{5}{2}$$)
f'(x) = $$\frac{1}{2 \sqrt{25-4 x^{2}}}$$ (-8x)
= –$$\frac{4x}{\sqrt{25-4 x^{2}}}$$
f(x) is increasing when f'(x) > 0
⇒ $$\frac{-4x}{\sqrt{25-4 x^{2}}}$$ > 0
i.e., x < o
f(x) is increasing when (-$$\frac{5}{2}$$, 0)
f(x) is decreasing when f'(x) < 0
⇒ –$$\frac{4x}{\sqrt{25-4 x^{2}}}$$ < 0
∴ x > 0
f(x) is decreasing when (0, $$\frac{5}{2}$$).

vii) ln (ln(x)); x > 1.
Solution:
f'(x) = –$$\frac{1}{lnx}•\frac{1}{x}$$
f(x) is decreasing when f'(x) > 0
$$\frac{1}{x.ln x}$$ >0
⇒ x. In x > 0
ln x is real only when x > 0
∴ ln x < 0 = ln 1
i.e., x > 1
f(x) is increasing when x > 1 i.e., in (1, ∞)
f(x) is decreasing when f'(x) < 0 ⇒ ln x > 0 = ln 1
i.e., x < 1
f(x) is decreasing in (0, 1)

viii) x³ + 3x² – 6x + 12.
Solution:
f(x) = x³ + 3x² – 6x + 12
f(x) = 3x² + 6x – 6
= 3(x² + 2x – 2)
= 3((x + 1)² – 3)
= 3[(x + 1) + √3] [(x + 1) – √3]
= 3(x + (1 + √3) (x + (1 – √3 )
f (x) < 0
⇒ x = -(1+ √3 ) or -(1 – √3 )
x = -1 – √3 or √3 – 1
f(x) is decreasing in (-1 -√3 , √3 -1)
f (x) > 0
f(x) increasing when (-∞, -1 – √3 ) ∪ (√3 – 1, ∞) Question 2.
Show that f(x) = cos²x is strictly increasing on (0, π/2).
Solution:
f(x) = cos² X
⇒ f(x) = 2 cos x (-sin x)
= -2 sin x cos x
= -sin 2x
Since 0 < x < $$\frac{\pi}{2}$$
⇒ 0 < 2x < π
Since ‘sin x’ is +ve between 0 and π
∴ f(x) is clearly -ve.
∴ f'(x) < 0
∴ f(x) is strictly decreasing.

Question 3.
Show that x + $$\frac{1}{x}$$ is increasing on [1, ∞)
Solution:
Let f(x) = x + $$\frac{1}{x}$$
f'(x) = 1 – $$\frac{1}{x^{2}}$$ = $$\frac{x^{2}-1}{x^{2}}$$
Since x ∈ [1, ∞) = $$\frac{x^{2}-1}{x^{2}}$$ > 0
∴ f'(x) > 0
∴ f(x) is increasing.

Question 4.
Show that $$\frac{x}{1+x}$$ < ln (1 + x) < x ∀ x > 0
Solution:
Let f(x) = ln(1 + x)- $$\frac{x}{1+x}$$
= ln(1 + x) – $$\frac{1+x-1}{1+x}$$
= ln(1 + x) – 1 + $$\frac{1}{1+x}$$
f'(x) = $$\frac{1}{1+x}$$ – $$\frac{1}{(1+x)^{2}}$$
= $$\frac{1+x-1}{(1+x)^{2}}$$
= $$\frac{x}{(1+x)^{2}}$$ > 0 since x > 0
f(x) is increasing when x > 0
∴ f(x) > f(0)
f(0) = ln 1 – $$\frac{0}{1+0}$$ = 0 – 0 = 0
Since xe [1, ∞) =
ln (1 + x) – $$\frac{x}{1+x}$$ > 0
⇒ ln (1 + x) > $$\frac{x}{1+x}$$ ……… (1)
Let g(x) = x – ln (1 + x)
g'(x) = 1 – $$\frac{x}{1+x}=\frac{1+x-1}{1+x}$$
= $$\frac{x}{1+x}$$ > 0 since x > 0
g(x) is increasing when x > 0
i.e., g(x) > g(0)
g(0) = 0 – ln (1) = 0 – 0 = 0
∴ x – ln (1 + x) > 0
x > ln(1 + x) ………….. (2)
From (1), (2) we get
$$\frac{x}{1+x}$$ < ln (1 + x) < x ∀ x > 0

III.

Question 1.
Show that $$\frac{x}{1+x^{2}}$$ < tan-1 x < x when x > 0.
Solution:
Let f(x) = tan-1 x – $$\frac{x}{1+x^{2}}$$ f(x) is increasing when x > 0
f(x) > f(0)
But f(0) = tan-1 0 – 0 = 0 – 0 = 0
i.e., f(x) > 0 g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – tan-1 0 = 0 – 0 = 0
∴ x – tan-1 x > 0
⇒ x > tan-1 x ………. (2)
From (1), (2) we get
$$\frac{x}{1+x^{2}}$$ <tan-1 x< x for x > 0

Question 2.
Show that tan x > x for all (0, $$\frac{\pi}{2}$$)
Solution:
Let f(x) = tan x – x
f'(x) = sec² x – 1 > 0 for every
x ∈ (0, $$\frac{\pi}{2}$$)
f(x) is increasing for every x ∈ (0, $$\frac{\pi}{2}$$)
i.e., f(x) > f(0)
f(0) = tan 0 – 0 = 0 – 0 = 0
∴ tan x – x > 0
⇒ tan x > x for every x ∈ (0, $$\frac{\pi}{2}$$) Question 3.
If x ∈ (0, $$\frac{\pi}{2}$$) then show that $$\frac{2x}{\pi}$$ < sin x < x.
Solution:
Let f(x) = x – sin x
f(x) = 1- cos x > 0 for every x ∈ (0, $$\frac{\pi}{2}$$)
f(x) is increasing for every x ∈ (0, $$\frac{\pi}{2}$$)
⇒ f(x) > f(0)
f(0) = 0 – sin 0 = 0 – 0 = 0
∴ x – sin x > 0
⇒ x > sin x ………….. (1)
Let g(x) = sin x – $$\frac{2x}{\pi}$$
g'(x) = cos x – $$\frac{2}{\pi}$$ > 0 for every x ∈ (0, $$\frac{\pi}{2}$$)
g(x) is increasing in (0, $$\frac{\pi}{2}$$)
g(x) > g(0)
g(0) = sin 0 – 0 = 0 – 0 = 0
∴ sin x – $$\frac{2x}{\pi}$$ > 0
⇒ sin x > $$\frac{2x}{\pi}$$ ………… (2)
From (1), (2) we get
$$\frac{2x}{\pi}$$ < sin x < x for every x ∈ (0, $$\frac{\pi}{2}$$)

Question 4.
If x e (0,1) then show that 2x < ln $\frac{(1+x)}{(x-1)}$ < 2x [1 + $$\frac{x^{2}}{2(1+x^{2})}$$] Solution:
Let f(x) = ln $$\frac{(1+x)}{1-x}$$ – 2x
= ln (1 + x) – ln (1 – x) – 2x f(x) is increasing ih (0, 1)
i.e., x > 0 ⇒ f(x) > f(0)
f(0) = ln 1 – 0 = 0 – 0 = 0  g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – ln 1 = 0 – 0 = 0 for x ∈ (0,1)

Question 5.
At what point the slopes of the tangents y = $$\frac{x^{3}}{6}-\frac{3x^{3}}{2}+\frac{11x}{2}$$ + 12 increases?
Solution:
Equation of the curve is
y = $$\frac{x^{3}}{6}-\frac{3}{2}x^{2}+\frac{11x}{2}$$ + 12 Slope = m = $$\frac{x^{2}}{c}-3x+\frac{11}{2}$$
$$\frac{dm}{dx}$$ = $$\frac{2x}{2}$$ -3 = x – 3
Slope increases ⇒ m > 0
x – 3 > 0
x > 3
The slope increases in (3, ∝)

Question 6.
Show that the functions ln $$\frac{(1+x)}{x}$$ and $$\frac{x}{(1+x)ln(1+x)}$$ are decrasing on (0, ∞).
Solution:
i)  ∴ f(x) is decreasing for x ∈ (0, ∝)

ii) let f(x) = $$\frac{x}{(1+x)ln(1+x)}$$ ∴ f(x) is decreasing for x ∈ (0, ∝)

Question 7.
Find the intervals in which the function f (x) = x3 – 3×2 + 4 is strictly increasing all x e R.
Solution:
f(x) = x³ – 3x² + 4
f'(x) = 3x² – 6x
f(x) is increasing if f'(x) > 0
3x² – 6x > 0
3x(x – 2) > 0
(x – 0)(x – 2) > 0
f(x) is increasing if x £ (-∞, 0) u (0, ∞)
f(x) is decreasing if f'(x) < 0
(x – 0) (x – 2) < 0
x ∈ (0, 2) Question 8.
Find the intervals in which the function f(x) = sin4x + cos4x ∀ x ∈ [0, $$\frac{\pi}{2}$$] is increasing and decreasing.
Solution:
f(x) = sin4x + cos4x
f(x) = (sin²x)² + (cos²x)²
= (sin²x + cos²x)² – 2sin²x cos²x
= 1 – $$\frac{1}{2}$$ sin² 2x
f'(x) = $$\frac{-1}{2}$$ 2sin 2x. cos 2x(2)
= -2 sin 2x. cos 2x
= -sin 4x
Let 0 < x < $$\frac{\pi}{4}$$
∴ f(x) is decreasing if f'(x) < 0
⇒ -sinx < 0 ⇒ sinx > 0
∴ x ∈ (0, $$\frac{\pi}{4}$$)
f(x) is increasing if f'(x) > 0
⇒ – sinx > 0
⇒ sinx < 0
∴ x ∈ ($$\frac{\pi}{4}$$, $$\frac{\pi}{2}$$)