In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 pdf, free NCERT solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 book pdf download.

Board | CBSE |

Textbook | NCERT |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 6 |

Chapter Name | Application of Derivatives |

Exercise | Ex 6.2 |

Number of Questions Solved | 19 |

Category | NCERT Solutions |

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

**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

(b) strictly decreasing in

(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

⇒ f(x) is strictly increasing on

(b) f’ (x) = cos x is a -ve in the interval

⇒ f (x) is strictly decreasing in

(c) f’ (x) = cos x is +ve in the interval

while f’ (x) is -ve in the interval

∴ 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 =

The point 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

For f (x) to be increasing f’ (x) > 0

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}

∴ = 4x^{3} – 12x^{2} + 8x

For the function to be increasing >0

4x^{3} – 12x^{2} + 8x>0

⇒ 4x(x – 1)(x – 2)>0

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

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

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

Prove that is an increasing function of θ in **Solution:**

For the function to be increasing > 0

⇒ cosθ(4-cos^{2}θ)>0

⇒ cosθ>0

⇒ θ∈

**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) = ; When takes the

values x > 0, > 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

(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

∴ 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)

(c)

(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 is strictly increasing on I.**Solution:**

Given

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 and strictly decreasing on**Solution:**

f’ (x) =

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

When < 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 and strictly increasing on **Solution:**

f’ (x) =

In the interval ,f’ (x) = -ve

∴ f is strictly decreasing.

In the interval , 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, ∞)

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