# NCERT Solutions For Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 6 |

Exercise | Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2 |

Number of Questions Solved | 19 |

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

**NCERT TEXTBOOK EXERCISES**

**Ex 6.2 Class 12 Maths Question 1.**

Show that the function given by f (x) = 3x+17 is strictly increasing on R.

**Solution:**

f(x) = 3x + 17

∴ f’ (x) = 3>0 ∀ x∈R

⇒ f is strictly increasing on R.

**Ex 6.2 Class 12 Maths Question 2.**

Show that the function given by f (x) = e^{2x} is strictly increasing on R.

**Solution:**

We have f (x) = e^{2x}

⇒ f’ (x) = 2e^{2x}

Case I When x > 0, then f’ (x) = 2e^{2x}

**Ex 6.2 Class 12 Maths Question 3.**

Show that the function given by f (x) = sin x is

(a) strictly increasing in $\left( 0,\frac { \pi }{ 2 } \right)$

(b) strictly decreasing in $\left( \frac { \pi }{ 2 } ,\pi \right)$

(c) neither increasing nor decreasing in (0, π)

**Solution:**

We have f(x) = sinx

∴ f’ (x) = cosx

(a) f’ (x) = cos x is + ve in the interval $\left( 0,\frac { \pi }{ 2 } \right)$

⇒ f(x) is strictly increasing on $\left( 0,\frac { \pi }{ 2 } \right)$

(b) f’ (x) = cos x is a -ve in the interval $\left( \frac { \pi }{ 2 } ,\pi \right)$

⇒ f (x) is strictly decreasing in $\left( \frac { \pi }{ 2 } ,\pi \right)$

(c) f’ (x) = cos x is +ve in the interval $\left( 0,\frac { \pi }{ 2 } \right)$

while f’ (x) is -ve in the interval $\left( \frac { \pi }{ 2 } ,\pi \right)$

∴ f(x) is neither increasing nor decreasing in (0,π)

**Ex 6.2 Class 12 Maths Question 4.**

Find the intervals in which the function f given by f(x) = 2x² – 3x is

(a) strictly increasing

(b) strictly decreasing

**Solution:**

f(x) = 2x² – 3x

⇒ f’ (x) = 4x – 3

⇒ f’ (x) = 0 at x = 34

The point x=34 divides the real

**Ex 6.2 Class 12 Maths Question 5.**

Find the intervals in which the function f given by f (x) = 2x^{3} – 3x² – 36x + 7 is

(a) strictly increasing

(b) strictly decreasing

**Solution:**

f(x) = 2x^{3} – 3x² – 36x + 7;

f (x) = 6 (x – 3) (x + 2)

⇒ f’ (x) = 0 at x = 3 and x = – 2

The points x = 3, x = – 2, divide the real line into three disjoint intervals viz. (-∞,-2), (-2,3), (3,∞)

Now f’ (x) is +ve in the intervals (-∞, -2) and (3,∞). Since in the interval (-∞, -2) each factor x – 3, x + 2 is -ve.

⇒ f’ (x) = + ve.

(a) f is strictly increasing in (-∞, -2)∪(3,∞)

(b) In the interval (-2,3), x+2 is +ve and x-3 is -ve.

f (x) = 6(x – 3)(x + 2) = + x – = -ve

∴ f is strictly decreasing in the interval (-2,3).

**Ex 6.2 Class 12 Maths Question 6.**

Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x² + 2x – 5

(b) 10 – 6x – 2x²

(c) – 2x^{3} – 9x² – 12x + 1

(d) 6 – 9x – x²

(e) (x + 1)^{3}(x – 3)^{3}

**Solution:**

(c) Let f(x) = – 2x^{3} – 9x^{2} – 12x + 1

∴ f’ (x) = – 6x^{2} – 18x – 12

= – 6(x^{2} + 3x + 2)

f'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2

The points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – ∞, – 2) ( – 2, – 1) and( – 1 ∞).

In the interval (-∞,-2) i.e.,-∞<x<-2 (x+ 1) (x+2) are -ve.

∴f’ (x) = (-) (-) (- ) = – ve.

⇒ f (x) is decreasing in (-∞,-2)

In the interval (-2, -1) i.e., – 2 < x < -1,

(x + 1) is -ve and (x + 2) is + ve.

∴ f'(x) = (-)(-) (+) = + ve.

⇒ f (x) is increasing in (-2, -1)

In the interval (-1,∞) i.e.,-1 <x<∞,(x + 1) and (x + 2) are both positive. f’ (x) = (-) (+) (+) = -ve.

⇒ f (x) is decreasing in (-1, ∞)

Hence, f (x) is increasing for – 2 < x < – 1 and decreasing for x<-2 and x>-1.

**Ex 6.2 Class 12 Maths Question 7.**

Show that $Y=log(1+x)-\frac { 2x }{ 2+x } x>-1$, is an increasing function of x throughout its domain.

**Solution:**

let $f(x)=log(1+x)-\frac { 2x }{ 2+x } x>-1$

f’ (x) = $\frac { { x }^{ 2 } }{ { (x+1)(x+2) }^{ 2 } }$

For f (x) to be increasing f’ (x) > 0

⇒ $\Rightarrow \frac { 1 }{ x+1 } >0\Rightarrow x>-1$

Hence, $y=log(1+x)-\frac { 2x }{ 2+x }$ is an increasing function of x for all values of x > – 1.

**Ex 6.2 Class 12 Maths Question 8.**

Find the values of x for which y = [x (x – 2)]² is an increasing function.

**Solution:**

y = x^{4} – 4x^{3} + 4x^{2}

∴ dy/dx = 4x^{3} – 12x^{2} + 8x

For the function to be increasing dy/dx >0

4x^{3} – 12x^{2} + 8x>0

⇒ 4x(x – 1)(x – 2)>0

For 0 < x < 1, dy/dx = (+)(-)(-) = +ve and for x > 2, dy/dx = (+) (+) (+) = +ve

Thus, the function is increasing for 0 < x < 1 and x > 2.

**Ex 6.2 Class 12 Maths Question 9.**

Prove that $y=\frac { 4sin\theta }{ (2+cos\theta ) } -\theta$ is an increasing function of θ in [0,π/2]

**Solution:**

$\frac { dy }{ dx } =\frac { 8cos\theta +4 }{ { (2+cos\theta ) }^{ 2 } } -1=\frac { cos\theta (4-cos\theta ) }{ { (2+cos\theta ) }^{ 2 } }$

For the function to be increasing dy/dx > 0

⇒ cosθ(4-cos^{2}θ)>0

⇒ cosθ>0

⇒ θ∈ $\left[ 0,\frac { \pi }{ 2 } \right]$1

**Ex 6.2 Class 12 Maths Question 10.**

Prove that the logarithmic function is strictly increasing on (0, ∞).

**Solution:**

Let f (x) = log x

Now, f’ (x) = 1/x ; When takes the

values x > 0, 1/x > 0, when x > 0,

∵ f’ (x) > 0

Hence, f (x) is an increasing function for x > 0 i.e

**Ex 6.2 Class 12 Maths Question 11.**

Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (-1,1).

**Solution:**

Given

f (x) = x² – x + 1

∴ f (x) is neither increasing nor decreasing on (-1,1).

**Ex 6.2 Class 12 Maths Question 12.**

Which of the following functions are strictly decreasing on $\left[ 0,\frac { \pi }{ 2 } \right]$

(a) cos x

(b) cos 2x

(c) cos 3x

(d) tan x

**Solution:**

(a) We have f (x) = cos x

∴ f’ (x) = – sin x < 0 in $\left[ 0,\frac { \pi }{ 2 } \right]$

∴ f’ (x) is a decreasing function.

**Ex 6.2 Class 12 Maths Question 13.**

On which of the following intervals is the function f given by f (x )= x^{100} + sin x – 1 strictly decreasing ?

(a) (0,1)

(b) $\left[ \frac { \pi }{ 2 } ,\pi \right]$

(c) $\left[ 0,\frac { \pi }{ 2 } \right]$

(d) none of these

**Solution:**

(d) f(x) = x^{100} + sin x – 1

∴ f’ (x)= 100x^{99}+ cos x

(a) for(-1, 1)i.e.,- 1 <x< 1,-1 <x^{99}< 1

⇒ -100<100x^{99}<100;

Also 0 ⇒ f’ (x) can either be +ve or -ve on(-1, 1)

∴ f (x) is neither increasing nor decreasing on (-1,1).

(b) for (0,1) i.e. 0<x< 1 x^{99} and cos x are both +ve ∴ f’ (x) > 0

⇒ f (x) is increasing on(0,1)

**Ex 6.2 Class 12 Maths Question 14.**

Find the least value of a such that the function f given by f (x) = x² + ax + 1 is strictly increasing on (1,2).

**Solution:**

We have f (x) = x² + ax + 1

∴ f’ (x) = 2x + a.

Since f (x) is an increasing function on (1,2)

f’ (x) > 0 for all 1 < x < 2 Now, f” (x) = 2 for all x ∈ (1,2) ⇒ f” (x) > 0 for all x ∈ (1,2)

⇒ f’ (x) is an increasing function on (1,2)

⇒ f’ (x) is the least value of f’ (x) on (1,2)

But f’ (x)>0 ∀ x∈ (1,2)

∴ f’ (1)>0 =>2 + a>0

⇒ a > – 2 : Thus, the least value of a is – 2.

**Ex 6.2 Class 12 Maths Question 15.**

Let I be any interval disjoint from (-1,1). Prove that the function f given by $f(x)=x+\frac { 1 }{ x }$ is strictly increasing on I.

**Solution:**

Given

$f(x)=x+\frac { 1 }{ x }$

Hence, f’ (x) is strictly increasing on I.

**Ex 6.2 Class 12 Maths Question 16.**

Prove that the function f given by f (x) = log sin x is strictly increasing on (0,π/2) and strictly decreasing on

(π/2,π)

**Solution:**

f’ (x) = $\frac { 1 }{ sin\quad x } .cos\quad x\quad cot\quad x\quad$

when 0 < x < π/2, f’ (x) is +ve; i.e., increasing

When π/2 < x < π, f’ (x) is – ve; i.e., decreasing,

∴ f (x) is decreasing. Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).

**Ex 6.2 Class 12 Maths Question 17.**

Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, π/2) and strictly increasing on (π/2, π)

**Solution:**

f(x)= log cosx

f’ (x) = $\frac { 1 }{ cosx } (-sinx)=-tanx$

In the interval (0,π/2) ,f’ (x) = -ve

∴ f is strictly decreasing.

In the interval (π/2,π), f’ (x) is + ve.

∴ f is strictly increasing in the interval.

**Ex 6.2 Class 12 Maths Question 18.**

Prove that the function given by

f (x) = x^{3} – 3x^{2} + 3x -100 is increasing in R.

**Solution:**

f’ (x) = 3x^{2} – 6x + 3

= 3 (x^{2} – 2x + 1)

= 3 (x -1 )^{2}

Now x ∈ R, f'(x) = (x – 1)^{2}≥0

i.e. f'(x)≥0 ∀ x∈R; hence, f(x) is increasing on R.

**Ex 6.2 Class 12 Maths Question 19.**

The interval in which y = x^{2} e^{-x} is increasing is

(a) (-∞,∞)

(b) (-2,0)

(c) (2,∞)

(d) (0,2)

**Solution:**

(d) f’ (x) = 2xe^{-x }+ x^{2}( – e^{-x}) = xe^{-x}(2-x) = e^{-x}x(2-x)

Now e^{-x} is positive for all x ∈ R f’ (x) = 0 at x = 0,2

x = 0, x = 2 divide the number line into three disjoint intervals, viz. (-∞, 0), (0,2), (2, ∞)

(a) Interval (-∞,0) x is +ve and (2-x) is +ve

∴ f’ (x) = e^{-x}x (2- x)=(+)(-) (+) = -ve

⇒ f is decreasing in (-∞,0)

(b) Interval (0,2) f’ (x) = e^{-x} x (2 – x)

= (+)(+)(+) = +ve

⇒ f is increasing in (0,2)

(c) Interval (2, ∞) f’ (x) = e^{-x} x (2 – x) = (+) (+) (-)

= – ve

⇒ f is decreasing in the interval (2, ∞)

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2 PDF

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