# NCERT Solutions for Class 12 Maths Application of Derivatives

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**NCERT Solutions for Class 12 Maths Application of Derivative Exercise 6.2**## NCERT Solutions for Class 12 Maths Application of Derivatives Exercise 6.2

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 1.**

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

**NCERT Solutions:**

We are given that f(x) = 3x + 17.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get f ‘(x) = 3 > 0 for all x ∊ R.

Thus, f(x) = 3x + 17 is an strictly increasing function on R.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 2.**

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

**NCERT Solutions:**

We are given that f(x) = e^{2x}.

Since f(x) is an exponential function, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get f ‘(x) = e^{2x} . 2 = 2e^{2x} > 0 for all x ∊ R.

Thus, f(x) = e^{2x} is an strictly increasing function on R.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 3.**

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

(a) strictly increasing in (0, π/2)

(b) strictly decreasing in (π/2, π)

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

**NCERT Solutions:**

We are given that f(x) = sin x.

Since f(x) is a sine function, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get f ‘(x) = cos x.

(a) In x ∊ (0, π/2), cos x > 0 ⇒ f ‘(x) > 0

Thus, f(x) = sin x is an strictly increasing function on x ∊ (0, π/2).

(b) In x ∊ (π/2, π), cos x < 0 ⇒ f ‘(x) < 0

Thus, f(x) = sin x is an strictly decreasing function on x ∊ (π/2, π).

(c) In x ∊ (0, π/2), we have f ‘(x) > 0 and in x ∊ (π/2, π), we have f ‘(x) < 0.

Thus, the given function f(x) = sin x is neither increasing nor decreasing in (0, π).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 4.**

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

(a) strictly increasing

(b) strictly decreasing

**NCERT Solutions:**

We are given that f(x) = 2x^{2} – 3x.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get f ‘(x) = 4x – 3.

(a) For strictly increasing f ‘(x) > 0 ⇒ 4x – 3 > 0 ⇒ x > 3/4 ⇒ x ∊ (3/4, ∞).

Thus, the given function f(x) = 2x^{2} – 3x is strictly increasing in x ∊ (3/4, ∞).

(b) For strictly decreasing f ‘(x) < 0 ⇒ 4x – 3 < 0 ⇒ x < 3/4 ⇒ x ∊ (-∞, 3/4).

Thus, the given function f(x) = 2x^{2} – 3x is strictly decreasing in x ∊ (-∞, 3/4).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 5.**

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

(a) strictly increasing

(b) strictly decreasing

**NCERT Solutions:**

We are given that f(x) = 2x^{3} – 3x^{2} – 36x + 7.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = 6x^{2} – 6x – 36 = 6(x^{2} – x – 6) = 6(x – 3)(x + 2).

(a) For strictly increasing f ‘(x) > 0 ⇒ 6(x – 3)(x + 2) > 0 ⇒ x < -2 OR x > 3 ⇒ x ∊ (-∞, -2) U (3, ∞).

Thus, the given function f(x) = 2x^{3} – 3x^{2} – 36x + 7 is strictly increasing in x ∊ (-∞, -2) U (3, ∞).

(b) For strictly decreasing f ‘(x) < 0 ⇒ 6(x – 3)(x + 2) < 0 ⇒ -2 < x < 3 ⇒ x ∊ (-2, 3).

Thus, the given function f(x) = 2x^{3} – 3x^{2} – 36x + 7 is strictly decreasing in x ∊ (-2, 3).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 6.**

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

(a) x^{2} + 2x – 5

**NCERT Solutions:**

We are given that f(x) = x^{2} + 2x – 5.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = 2x + 2 = 2(x + 1).

For strictly increasing f ‘(x) > 0 ⇒ 2(x + 1) > 0 ⇒ x > -1 ⇒ x ∊ (-1, ∞).

Thus, the given function f(x) = x^{2} + 2x – 5 is strictly increasing in x ∊ (-1, ∞).

For strictly decreasing f ‘(x) < 0 ⇒ 2(x + 1) < 0 ⇒ x < -1 ⇒ x ∊ (-∞, -1).

Thus, the given function f(x) = x^{2} + 2x – 5 is strictly decreasing in x ∊ (-∞, -1).

(b) 10 – 6x – 2x^{2}

**NCERT Solutions:**

We are given that f(x) = 10 – 6x – 2x^{2}.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = -6 – 4x = -4(x + 3/2).

For strictly increasing f ‘(x) > 0 ⇒ -4(x + 3/2) > 0 ⇒ (x + 3/2) < 0 ⇒ x < -3/2 ⇒ x ∊ (-∞, -3/2).

Thus, the given function f(x) = 10 – 6x – 2x^{2} is strictly increasing in x ∊ (-∞, -3/2).

For strictly decreasing f ‘(x) < 0 ⇒ -4(x + 3/2) < 0 ⇒ (x + 3/2) > 0 ⇒ x > -3/2 ⇒ x ∊ (-3/2, ∞).

Thus, the given function f(x) = 10 – 6x – 2x^{2} is strictly decreasing in x ∊ (-3/2, ∞).

(c) –2x^{3} – 9x^{2} – 12x + 1

**NCERT Solutions:**

We are given that f(x) = –2x^{3} – 9x^{2} – 12x + 1.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = -6x^{2} – 18x – 12 = -6(x^{2} + 3x + 2) = -6(x + 2)(x + 1).

For strictly increasing

f ‘(x) > 0 ⇒ -6(x + 2)(x + 1) > 0 ⇒ (x + 2)(x + 1) < 0 ⇒ -2 < x < -1 ⇒ x ∊ (-2, -1).

Thus, the given function f(x) = –2x^{3} – 9x^{2} – 12x + 1 is strictly increasing in x ∊ (-2, -1).

For strictly decreasing

f ‘(x) > 0 ⇒ -6(x + 2)(x + 1) < 0 ⇒ (x + 2)(x + 1) > 0 ⇒ x < -2 OR x > -1 ⇒ x ∊ (-∞, -2) U (-1, ∞).

Thus, the given function f(x) = –2x^{3} – 9x^{2} – 12x + 1 is strictly decreasing in x ∊ (-∞, -2) U (-1, ∞).

(d) 6 – 9x – x^{2}

**NCERT Solutions:**

We are given that f(x) = 6 – 9x – x^{2}.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = -9 – 2x = -2(x + 9/2).

For strictly increasing

f ‘(x) > 0 ⇒ -2(x + 9/2) > 0 ⇒ (x + 9/2) < 0 ⇒ x < -9/2 ⇒ x ∊ (-∞, -9/2).

Thus, the given function f(x) = 6 – 9x – x^{2} is strictly increasing in x ∊ (-∞, -9/2).

For strictly decreasing

f ‘(x) < 0 ⇒ -2(x + 9/2) < 0 ⇒ (x + 9/2) > 0 ⇒ x > -9/2 ⇒ x ∊ (-9/2, ∞).

Thus, the given function f(x) = 6 – 9x – x^{2} is strictly decreasing in x ∊ (-9/2, ∞).

(e) (x + 1)^{3}(x – 3)^{3}

**NCERT Solutions:**

We are given that f(x) = (x + 1)^{3}(x – 3)^{3}.

Since f(x) is a polynomial, it is continuous and differential in R.

On differentiating f(x) with respect to x, we get

f ‘(x) = 3(x + 1)^{2} . (x – 3)^{3} + (x + 1)^{3 }.3(x – 3)^{2 }

= 3(x + 1)^{2}(x – 3)^{2}(x – 3 + x + 1)

= 3(x + 1)^{2}(x – 3)^{2}(2x – 2)

= 6(x + 1)^{2}(x – 3)^{2}(x – 1)

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 7.**

Show that y = log(1 + x) – 2x/(2 + x), x > – 1, is an increasing function of x throughout its domain.

**NCERT Solutions:**

We are given that

On differentiating with respect to x, we get

Thus, the given function y = log(1 + x) – 2x/(2 + x), x > – 1, is an increasing function of x throughout its domain.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 8.**

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

**NCERT Solutions:**

We are given that y = [x(x – 2)]^{2} = (x^{2} – 2x)^{2}.

On differentiating with respect to x, we get

For increasing or decreasing, put dy/dx = 0

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 9.**

Prove that y = 4sin θ / (2 + cos θ) – θ is an increasing function of θ in [0, π2].

**NCERT Solutions:**

We are given that y = 4sin θ / (2 + cos θ) – θ.

On differentiating with respect to x, we get

Thus, the given function y = 4sin θ / (2 + cos θ) – θ is increasing function of θ in [0, π2].

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 10.**

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

**NCERT Solutions:**

We are given that y = log x. Its domain is (0, ∞).

On differentiating with respect to x, we get

Thus, the logarithmic function y = log x is increasing function of θ in (0, ∞).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 11.**

Prove that the function f given by f (x) = x^{2} – x + 1 is neither strictly increasing nor strictly decreasing on (– 1, 1).

**NCERT Solutions:**

We are given that y = f(x) = x^{2} – x + 1.

On differentiating with respect to x, we get

f ‘(x) = 2x – 1

For increasing, f ‘(x) > 0 ⇒ 2x – 1 > 0 ⇒ x > 1/2.

For decreasing, f ‘(x) < 0 ⇒ 2x – 1 < 0 ⇒ x < 1/2.

Thus, the given function is neither strictly increasing nor strictly decreasing on (– 1, 1).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 12.**

Which of the following functions are strictly decreasing on (0, π/2)?

(A) cos x

(B) cos 2x

(C) cos 3x

(D) tan x

**NCERT Solutions:**

(A)

We are given that y = f(x) = cos x.

On differentiating with respect to x, we get

f ‘(x) = -sin x

Since f ‘(x) = -sin x < 0 for all x ∊ (0, π/2).

Thus, the given function is strictly decreasing on (0, π/2).

(B)

We are given that y = f(x) = cos 2x.

On differentiating with respect to x, we get

f ‘(x) = -2sin 2x

Since f ‘(x) = -2sin 2x < 0 for all x ∊ (0, π/2). As

Thus, the given function is strictly decreasing on (0, π/2).

Therefore, the functions in (A) and (B) are strictly decreasing in (0, π/2).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 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) (π/2, π)

(C) (0, π/2)

(D) None of these

**NCERT Solutions:**

We are given that y = f(x) = x^{100} + sin x –1.

On differentiating with respect to x, we get

Since the given function is NOT decreasing in any of the intervals, hence the correct answer is D.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 14.**

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

**NCERT Solutions:**

We are given that y = f(x) = x^{2} + ax + 1.

On differentiating with respect to x, we get f ‘(x) = 2x + a.

Given that 1 < x < 2 ⇒ 2 < 2x < 4 ⇒ (2 + a) < 2x + a < (4 + a) ⇒ (2 + a) < f ‘(x) < (4 + a).

Now f(x) is strictly increasing on (1, 2) then f ‘(x) > 0 for 1 < x < 2.

⇒ (2 + a) ≧ 0 ⇒ a ≧ -2. Thus, required least value of a is -2.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 15.**

Let I be any interval disjoint from (–1, 1). Prove that the function f given by f(x) = x + 1/x is strictly increasing on I.

**NCERT Solutions:**

We are given that y = f(x) = x + 1/x.

On differentiating with respect to x, we get

Since I is an interval which is a subset of R – (-1, 1).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 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).

**NCERT Solutions:**

We are given that y = f(x) = log sin x.

On differentiating with respect to x, we get

Since cot x > 0 for all x ∊ (0, π/2) and cot x < 0 for all x ∊ (π/2, π), then f (x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 17.**

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

**NCERT Solutions:**

We are given that y = f(x) = log cos x.

On differentiating with respect to x, we get

As tan x > 0 for all x ∊ (0, π/2) ⇒ -tan x < 0 for all x ∊ (0, π/2) and

tan x < 0 for all x ∊ (π/2, π) ⇒ -tan x > 0 for all x ∊ (π/2, π).

Then f (x) = log cos x is strictly decreasing on (0, π/2) and strictly increasing on (π/2, π).

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 18.**

Prove that the function given by f (x) = x^{3} – 3x^{2} + 3x – 100 is increasing in R.

**NCERT Solutions:**

We are given that y = f(x) = log sin x.

On differentiating with respect to x, we get

f ‘(x) = 3x^{2} – 6x + 3 = 3(x^{2} – 2x + 1) = 3(x – 1)^{2} , (a perfect square) ⇒ f ‘(x) ≧ 0

Thus, the given function is increasing in R.

**NCERT Solutions for Class 12 Maths Application of Derivatives ****Exercise**** 6.2: Ques No 19.**

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

(A) (– ∞, ∞)

(B) (– 2, 0)

(C) (2, ∞)

(D) (0, 2)

**NCERT Solutions:**

We are given that y = f(x) = x^{2}e^{–x}.

On differentiating with respect to x, we get

For increasing, put dy/dx = 0