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.

Withotit using the derivative, show that

i) The function f(x) = 3x + 7 is strictly increasing on R.

Solution:

Let x_{1}, x_{2} ∈ R with x_{1} < x_{2}

Then 3x_{1} < 3x_{2}

Adding 7 on bothside

3x_{1} + 7 < 3x_{2} + 7

⇒ f(x_{1}) < f(x_{2})

∴ x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1} x_{2} ∈ 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 x_{1}, x_{2} ∈ R

Such that x_{1} < x_{2}

⇒ (\(\frac{1}{2}\))^{x1} > (\(\frac{1}{2}\))^{x2}

⇒ f(x_{1}) > f(x_{2})

∴ f(x) is strictly decreasing on R.

iii) The function f(x) = e^{3x} is strictly increasing on R.

Solution:

f(x) = e^{3x}

Let x_{1}, x_{2} ∈ R such that x_{1} < x_{2}

We know that of a > b then e^{a} > e^{b}

Then e^{3x} < e^{3x2}

⇒ f(x_{1}) < f(x_{2})

∴ f is strictly increasing on R.

iv) The function f(x) = 5. – 7x is strictly decreasing on R.

Solution:

f(x) = 5 – 7x

Let x_{1} x_{2} ∈ R

Such that x_{1} < x_{2}

Then 7x_{1} < 7x_{2}

-7x_{1} > -7x_{2}

Adding 5 on bothsides

5 – 7x_{1} > 5 – 7x_{2}

f(x_{1}) > (x_{2})

∴ x_{1} < x_{2} ⇒ f(x_{1}) > f(x_{2}) V x_{1} x_{2} ∈ 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

x^{6} – 1 – 3x^{4} + 3x ²

f'(x) = 6x^{5} – 12x³ + 6x

= 6(x^{5} – 2x³ + x)

= 6x(x^{4} – 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) xe^{x}

Solution:

f'(x) = x . e^{x} + e^{x}. 1 = e^{x}(x + 1)

e^{x} 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 [latex]\frac{(1+x)}{(x-1)}[/latex] < 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) = sin^{4}x + cos^{4}x ∀ x ∈ [0, \(\frac{\pi}{2}\)] is increasing and decreasing.

Solution:

f(x) = sin^{4}x + cos^{4}x

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 ∈ (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))