Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.2 | Set 1

Last Updated : 02 Jun, 2021

Question 1. Find the intervals in which the following functions are increasing or decreasing.

(i) f(x) = 10 – 6x – 2x²

Solution:

We are given,

f(x) = 10 – 6x – 2x²

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(10 - 6x - 2x^2)$

f'(x) = 0 – 6 – 4x

f'(x) = – 6 – 4x

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> – 6 – 4x > 0

=> – 4x > 6

=> x < –6/4

=> x < –3/2

=> x ∈ (–∞, –3/2)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> – 6 – 4x < 0

=> – 4x < 6

=> x > –6/4

=> x > –3/2

=> x ∈ ( –3/2, ∞)

Thus, f(x) is increasing on the interval x ∈ (–∞, –3/2) and decreasing on the interval x ∈ ( –3/2, ∞).

(ii) f(x) = x² + 2x – 5

Solution:

We are given,

f(x) = x² + 2x – 5

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^2 + 2x - 5)$

f'(x) = 2x + 2 – 0

f'(x) = 2x + 2

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> 2x + 2 > 0

=> 2x > –2

=> x > –2/2

=> x > –1

=> x ∈ (–1, ∞)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> 2x + 2 < 0

=> 2x < –2

=> x < –2/2

=> x < –1

=> x ∈ (–∞, –1)

Thus, f(x) is increasing on the interval x ∈ (–1, ∞) and decreasing on the interval x ∈ ( –∞, –1).

(iii) f(x) = 6 – 9x – x²

Solution:

We are given,

f(x) = 6 – 9x – x²

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(6 - 9x - x^2)$

f'(x) = 0 – 9 – 2x

f'(x) = – 9 – 2x

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> –9 – 2x > 0

=> –2x > 9

=> x > –9/2

=> x ∈ (–9/2, ∞)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> –9 – 2x < 0

=> –2x < 9

=> x < –9/2

=> x ∈ (–∞, –9/2)

Thus, f(x) is increasing on the interval x ∈ (–9/2, ∞) and decreasing on the interval x ∈ ( –∞, –9/2).

(iv) f(x) = 2x³ – 12x² + 18x + 15

Solution:

We are given,

f(x) = 2x³ – 12x²+ 18x + 15

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(2x^3 - 12x^2 + 18x + 15)$

f'(x) = 6x² – 24x + 18 + 0

f'(x) = 6x² – 24x + 18

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² – 24x + 18 = 0

=> 6 (x² – 4x + 3) = 0

=> x² – 4x + 3 = 0

=> x² – 3x – x + 3 = 0

=> x (x – 3) – 1 (x – 3) = 0

=> (x – 1) (x – 3) = 0

=> x = 1, 3

Clearly, f'(x) > 0 if x < 1 and x > 3.

Also, f'(x) < 0, if 1 < x < 3.

Thus, f(x) is increasing on the interval x ∈ (–∞, 1)∪ (3, ∞) and decreasing on the interval x ∈ (1, 3).

(v) f(x) = 5 + 36x + 3x² – 2x³

Solution:

We are given,

f(x) = 5 + 36x + 3x² – 2x³

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(5 + 36x + 3x^2 - 2x^3)$

f'(x) = 0 + 36 + 6x – 6x²

f'(x) = 36 + 6x – 6x²

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 36 + 6x – 6x² = 0

=> 6 (– x² + x + 6) = 0

=> 6 (–x² + 3x – 2x + 6) = 0

=> –x² + 3x – 2x + 6 = 0

=> x² – 3x + 2x – 6 = 0

=> (x – 3) (x + 2) = 0

=> x = 3, – 2

Clearly, f’(x) > 0 if –2 < x < 3.

Also f’(x) < 0 if x < –2 and x > 3.

Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (3, ∞).

(vi) f(x) = 8 + 36x + 3x² – 2x³

Solution:

We are given,

f(x) = 8 + 36x + 3x² – 2x³

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(8 + 36x + 3x^2 - 2x^3)$

f'(x) = 0 + 36 + 6x – 6x²

f'(x) = 36 + 6x – 6x²

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 36 + 6x – 6x² = 0

=> 6 (– x² + x + 6) = 0

=> 6 (–x² + 3x – 2x + 6) = 0

=> –x² + 3x – 2x + 6 = 0

=> x² – 3x + 2x – 6 = 0

=> (x – 3) (x + 2) = 0

=> x = 3, –2

Clearly, f’(x) > 0 if –2 < x < 3.

Also f’(x) < 0 if x < –2 and x > 3.

Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (3, ∞).

(vii) f(x) = 5x³ – 15x² – 120x + 3

Solution:

We are given,

f(x) = 5x³ – 15x² – 120x + 3

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(5x^3 - 15x^2 - 120x + 3)$

f'(x) = 15x²– 30x – 120 + 0

f'(x) = 15x² – 30x – 120

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 15x² – 30x – 120 = 0

=> 15(x² – 2x – 8) = 0

=> 15(x² – 4x + 2x – 8) = 0

=> x² – 4x + 2x – 8 = 0

=> (x – 4) (x + 2) = 0

=> x = 4, –2

Clearly, f’(x) > 0 if x < –2 and x > 4.

Also f’(x) < 0 if –2 < x < 4.

Thus, f(x) is increasing on x ∈ (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4).

(viii) f(x) = x³ – 6x²– 36x + 2

Solution:

We are given,

f(x) = x³ – 6x² – 36x + 2

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^3 - 6x^2 - 36x + 2)$

f'(x) = 3x² – 12x – 36 + 0

f'(x) = 3x² – 12x – 36

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x² – 12x – 36 = 0

=> 3(x² – 4x – 12) = 0

=> 3(x² – 6x + 2x – 12) = 0

=> x² – 6x + 2x – 12 = 0

=> (x – 6) (x + 2) = 0

=> x = 6, –2

Clearly, f’(x) > 0 if x < –2 and x > 6.

Also f’(x) < 0 if –2< x < 6

Thus, f(x) is increasing on x ∈ (–∞,–2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (–2, 6).

(ix) f(x) = 2x³ – 15x² + 36x + 1

Solution:

We are given,

f(x) = 2x³ – 15x² + 36x + 1

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)$

f'(x) = 6x²– 30x + 36 + 0

f'(x) = 6x² – 30x + 36

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² – 30x + 36 = 0

=> 6 (x² – 5x + 6) = 0

=> 6(x² – 3x – 2x + 6) = 0

=> x² – 3x – 2x + 6 = 0

=> (x – 3) (x – 2) = 0

=> x = 3, 2

Clearly, f’(x) > 0 if x < 2 and x > 3.

Also f’(x) < 0 if 2 < x < 3.

Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3).

(x) f(x) = 2x³ + 9x² + 12x + 1

Solution:

We are given,

f(x) = 2x³ + 9x² + 12x + 1

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(2x^3 + 9x^2 + 12x + 1)$

f'(x) = 6x² + 18x + 12 + 0

f'(x) = 6x² + 18x + 12

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² + 18x + 12 = 0

=> 6 (x² + 3x + 2) = 0

=> 6(x² + 2x + x + 2) = 0

=> x² + 2x + x + 2 = 0

=> (x + 2) (x + 1) = 0

=> x = –1, –2

Clearly, f’(x) > 0 if –2 < x < –1.

Also f’(x) < 0 if x < –1 and x > –2.

Thus, f(x) is increasing on x ∈ (–2,–1) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (–2, ∞).

(xi) f(x) = 2x³ – 9x² + 12x – 5

Solution:

We are given,

f(x) = 2x³ – 9x² + 12x – 5

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}( 2x^3 - 9x^2 + 12x - 5)$

f'(x) = 6x² – 18x + 12 – 0

f'(x) = 6x² – 18x + 12

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² – 18x + 12 = 0

=> 6 (x² – 3x + 2) = 0

=> 6(x² – 2x – x + 2) = 0

=> x² – 2x – x + 2 = 0

=> (x – 2) (x – 1) = 0

=> x = 1, 2

Clearly, f’(x) > 0 if x < 1 and x > 2.

Also f’(x) < 0 if 1 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (1, 2).

(xii) f(x) = 6 + 12x + 3x² – 2x³

Solution:

We are given,

f(x) = 6 + 12x + 3x² – 2x³

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(6 + 12x + 3x^2 - 2x^3)$

f'(x) = 0 + 12 + 6x – 6x²

f'(x) = 12 + 6x – 6x²

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 12 + 6x – 6x² = 0

=> 6 (–x² + x + 2) = 0

=> x² – x – 2 = 0

=> x² – 2x + x – 2 = 0

=> (x – 2) (x + 1) = 0

=> x = 2, –1

Clearly, f’(x) > 0 if –1 < x < 2.

Also f’(x) < 0 if x < –1 and x > 2.

Thus, f(x) is increasing on x ∈ (–1, 2) and f(x) is decreasing on interval x ∈ (–∞, –1) ∪ (2, ∞).

(xiii) f(x) = 2x³ – 24x + 107

Solution:

We are given,

f(x) = 2x³ – 24x + 107

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(2x^3 - 24x + 107)$

f'(x) = 6x² – 24 + 0

f'(x) = 6x²– 24

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² – 24 = 0

=> 6x² = 24

=> x² = 4

=> x = 2, –2

Clearly, f’(x) > 0 if x < –2 and x > 2.

Also f’(x) < 0 if –2 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).

(xiv) f(x) = –2x³ – 9x² – 12x + 1

Solution:

We are given,

f(x) = –2x³ – 9x² – 12x + 1

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1)$

f'(x) = –6x² – 18x – 12 + 0

f'(x) = –6x²– 18x – 12

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> –6x² – 18x – 12 = 0

=> 6 (–x² – 3x – 2) = 0

=> x² + 3x + 2 = 0

=> x² + 2x + x + 2 = 0

=> (x + 2) (x + 1) = 0

=> x = –2, –1

Clearly, f’(x) > 0 if x < –1 and x > –2.

Also, f’(x) < 0 if –2 < x < –1.

Thus, f(x) is increasing on x ∈ (–2, –1) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (–1, ∞).

(xv) f(x) = (x – 1) (x – 2)²

Solution:

We are given,

f(x) = (x – 1) (x – 2)²

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}((x - 1) (x - 2)^2)$

f'(x) = (x – 2)² + 2 (x – 1) (x – 2)

f'(x) = (x – 2) (x – 2 + 2x – 2)

f'(x) = (x – 2) (3x – 4)

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> (x – 2) (3x – 4) = 0

=> x = 2, 4/3

Clearly, f’(x) > 0 if x < 4/3 and x > 2.

Also, f’(x) < 0 if 4/3 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, 4/3) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (4/3, 2).

(xvi) f(x) = x³ – 12x² + 36x + 17

Solution:

We are given,

f(x) = x³ – 12x² + 36x + 17

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^3 - 12x^2 + 36x + 17)$

f'(x) = 3x² – 24x + 36 + 0

f'(x) = 3x² – 24x + 36

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x² – 24x + 36 = 0

=> 3 (x² – 8x + 12) = 0

=> x² – 8x + 12 = 0

=> x² – 6x – 2x + 12 = 0

=> (x – 6) (x – 2) = 0

=> x = 6, 2

Clearly, f’(x) > 0 if x < 2 and x > 6.

Also, f’(x) < 0 if 2 < x < 6.

Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (2, 6).

(xvii) f(x) = 2x³ – 24x + 7

Solution:

We are given,

f(x) = 2x³ – 24x + 7

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(2x^3 - 24x + 7)$

f'(x) = 6x² – 24 + 0

f'(x) = 6x² – 24

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x² – 24 = 0

=> 6x² = 24

=> x² = 4

=> x = 2, –2

Clearly, f’(x) > 0 if x < –2 and x > 2.

Also f’(x) < 0 if –2 < x < 2.

Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).

(xviii) f(x) = 3x⁴/10 – 4x³/5 -3x² + 36x/5 + 11

Solution:

We are given,

f(x) = 3x⁴/10 – 4x³/5 -3x² + 36x/5 + 11

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11)$

f'(x) = 6x³/5 – 12x²/5 -3(2x) + 36/5

f'(x) = 6/5[(x – 1)(x + 2)(x – 3)]

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6/5[(x – 1)(x + 2)(x – 3)] = 0

=> x = 1, –2, 3

Clearly, f’(x) > 0 if –2 < x < 1 and if x > 3

Also f’(x) < 0 if 1 < x < 3.

Thus, f(x) is increasing on x ∈ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).

(xix) f(x) = x⁴ – 4x

Solution:

We are given,

f(x) = x⁴ – 4x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^4 - 4x)$

f'(x) = 4x³ – 4

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x³ – 4 = 0

=> 4 (x³ – 1) = 0

=> x³ – 1 = 0

=> x³ = 1

=> x = 1

Clearly, f’(x) > 0 if x > 1.

Also f’(x) < 0 if x < 1.

Thus, f(x) is increasing on x ∈ (1, ∞), and f(x) is decreasing on interval x ∈ (–∞, 1).

(xx) f(x) = x⁴/4 + 2/3x³ – 5/2x² – 6x + 7

Solution:

We have,

f(x) = x⁴/4 + 2/3x³ – 5/2x² – 6x + 7

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\frac{x^4}{4}+\frac{2}{3}x^3-\frac{5}{2}x^2-6x+7)$

f'(x) = 4x³/4 + 6x²/3 – 10x/2 – 6 + 0

f'(x) = x³ + 2x² – 5x – 6

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> x³ + 2x² – 5x – 6 = 0

=> (x + 1) (x + 3) (x – 2) = 0

=> x = –1, –3, 2

Clearly f'(x) > 0 if –3 < x < –1 and x > 2.

Also f'(x) < 0 if x < –3 and –1 < x < 2.

Thus, f(x) is increasing on x ∈ (–3, –1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (–1, 2).

(xxi) f(x) = x⁴– 4x³ + 4x² + 15

Solution:

We have,

f(x) = x⁴ – 4x³ + 4x² + 15

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^4 - 4x^3 + 4x^2 + 15)$

f'(x) = 4x³ – 12x² + 8x + 0

f'(x) = 4x³ – 12x² + 8x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x³ – 12x² + 8x = 0

=> 4x (x² – 3x + 2) = 0

=> 4x (x – 2) (x – 1) = 0

=> x = 0, 1, 2

Clearly f'(x) > 0 if 0 < x < 1 and x > 2.

Also f'(x) < 0 if x < 0 and 1 < x < 2.

Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).

(xxii) f(x) = $5x^{\frac{3}{2}}-3x^\frac{5}{2}$ , x > 0

Solution:

We have,

f(x) = $5x^{\frac{3}{2}}-3x^\frac{5}{2}$

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(5x^{\frac{3}{2}}-3x^\frac{5}{2})$

f'(x) = $\frac{15}{2}x^{\frac{1}{2}}-\frac{15}{2}x^{\frac{3}{2}}$

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> $\frac{15}{2}x^{\frac{1}{2}}-\frac{15}{2}x^{\frac{3}{2}}$ = 0

=> $x^{\frac{1}{2}}-x^{\frac{3}{2}}$ = 0

=> x^1/2(1 – x) = 0

=> x = 0, 1

Clearly f'(x) > 0 if 0 < x < 1.

Also f'(x) < 0 if x > 0.

Thus, f(x) is increasing on x ∈ (0, 1) and f(x) is decreasing on interval x ∈ (1, ∞).

(xxiii) f(x) = x⁸ + 6x²

Solution:

We have,

f(x) = x⁸ + 6x²

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^8 + 6x^2)$

f'(x) = 8x⁷ + 12x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 8x⁷ + 12x = 0

=> 4x (2x⁶ + 3) = 0

=> x = 0

Clearly f'(x) > 0 if x > 0.

Also f'(x) < 0 if x < 0.

Thus, f(x) is increasing on x ∈ (0, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0).

(xxiv) f(x) = x³ – 6x² + 9x + 15

Solution:

We are given,

f(x) = x³ – 6x² + 9x + 15

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^3 - 6x^2 + 9x + 15)$

f'(x) = 3x² – 12x + 9 + 0

f'(x) = 3x² – 12x + 9

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 3x² – 12x + 9 = 0

=> 3 (x² – 4x + 3) = 0

=> x² – 4x + 3 = 0

=> x² – 3x – x + 3 = 0

=> (x – 3) (x – 1) = 0

=> x = 3, 1

Clearly f'(x) > 0 if x < 1 and x > 3.

Also f'(x) < 0 if 1 < x < 3.

Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).

(xxv) f(x) = [x(x – 2)]²

Solution:

We are given,

f(x) = [x(x – 2)]²

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}[x(x - 2)]^2$

f'(x) = $\frac{d}{dx}(x^2-2x)^2$

f'(x) = 2 (x² – 2x) (2x – 2)

f'(x) = 4x (x – 2) (x – 1)

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 4x (x – 2) (x – 1) = 0

=> x = 0, 1, 2

Clearly f'(x) > 0 if 0 < x < 1 and x > 2.

Also f'(x) < 0 if x < 0 and 1< x < 2.

Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).

(xxvi) f(x) = 3x⁴ – 4x³ – 12x² + 5

Solution:

We are given,

f(x) = 3x⁴ – 4x³ – 12x²+ 5

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(3x^4 - 4x^3 - 12x^2 + 5)$

f'(x) = 12x³ – 12x² – 24x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 12x³ – 12x² – 24x = 0

=> 12x (x² – x – 2) = 0

=> 12x (x + 1) (x – 2) = 0

=> x = 0, –1, 2

Clearly f'(x) > 0 if –1 < x < 0 and x > 2.

Also f'(x) < 0 if x < –1 and 0< x < 2.

Thus, f(x) is increasing on x ∈ (–1, 0) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –1) ∪ (0, 2).

(xxvii) f(x) = 3x⁴/2 – 4x³ – 45x² + 51

Solution:

We have,

f(x) = 3x⁴/2 – 4x³ – 45x² + 51

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\frac{3}{2}x^4-4x^3-45x^2+51)$

f'(x) = 6x³ – 12x² – 90x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> 6x³ – 12x² – 90x = 0

=> 6x (x² – 2x – 15) = 0

=> 6x (x + 3) (x – 5) = 0

=> x = 0, –3, 5

Clearly f'(x) > 0 if –3 < x < 0 and x > 5.

Also f'(x) < 0 if x < –3 and 0< x < 5.

Thus, f(x) is increasing on x ∈ (–3, 0) ∪ (5, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (0, 5).

(xxvii) f(x) = $\log(2+x)-\frac{2x}{2+x}$

Solution:

We have,

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

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\log(2+x)-\frac{2x}{2+x})$

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

f'(x) = $\frac{1}{2+x}-\frac{4+2x-2x}{(2+x)^2}$

f'(x) = $\frac{1}{2+x}-\frac{4}{(2+x)^2}$

f'(x) = $\frac{2+x-4}{(2+x)^2}$

f'(x) = $\frac{x-2}{(2+x)^2}$

Clearly f'(x) > 0 if x > 2.

Also f'(x) < 0 if x < 2

Thus, f(x) is increasing on x ∈ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 2).

Question 2. Determine the values of x for which the function f(x) = x² – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x² – 6x + 9 where the normal is parallel to the line y = x + 5.

Solution:

Given f(x) = x² – 6x + 9

On differentiating both sides with respect to x, we get

=> f’(x) = 2x – 6

=> f’(x) = 2(x – 3)

For f(x), we need to find the critical point, so we get,

=> f’(x) = 0

=> 2(x – 3) = 0

=> (x – 3) = 0

=> x = 3

Clearly, f’(x) > 0 if x > 3.

Also f’(x) < 0 if x < 3.

Thus, f(x) is increasing on (3, ∞) and f(x) is decreasing on interval x ∈ (–∞, 3).

Equation of the given curve is f(x) = x² – 6x + 9.

Slope of this curve is given by,

=> m₁ = dy/dx

=> m₁ = 2x – 6

And slope of the line is y = x + 5

Slope of this curve is given by,

=> m₂ = dy/dx

=> m₂ = 1

Now according to the question,

=> m₁m₂ = –1

=> 2x – 6 = –1

=> 2x = 5

=> x = 5/2

Putting x = 5/2 in the curve y = x² – 6x + 9, we get,

=> y = (5/2)² – 6 (5/2) + 9

=> y = 25/4 – 15 + 9

=> y = 1/4

Therefore, the required coordinates are (5/2, 1/4).

Question 3. Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.

Solution:

We have,

f(x) = sin x – cos x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(sin x – cos x)

f'(x) = cos x + sin x

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> cos x + sin x = 0

=> 1 + tan x = 0

=> tan x = –1

=> x = 3π/4 , 7π/4

Clearly f'(x) > 0 if 0 < x < 3π/4 and 7π/4 < x < 2π.

Also f'(x) < 0 if 3π/4 < x < 7π/4.

Thus, f(x) is increasing on x ∈ (0, 3π/4) ∪ (7π/4, 2π) and f(x) is decreasing on interval x ∈ (3π/4, 7π/4).

Question 4. Show that f(x) = e^2x is increasing on R.

Solution:

We have,

=> f(x) = e^2x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(e^{2x})$

f'(x) = 2e^2x

For f(x) to be increasing, we must have

=> f’(x) > 0

=> 2e^2x > 0

=> e^2x > 0

Now we know, the value of e lies between 2 and 3. Therefore, f(x) will be always greater than zero.

Thus, f(x) is increasing on interval R.

Hence proved.

Question 5. Show that f(x) = e^1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Solution:

We have,

=> f(x) = e^1/x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(e^{\frac{1}{x}})$

f'(x) = -e^x/x²

As x ∈ R, we have,

=> e^x > 0

Also, we get,

=> 1/x² > 0

This means, e^x/x² > 0

=> -e^x/x² < 0

Thus, f(x) is a decreasing function for all x ≠ 0.

Hence proved.

Question 6. Show that f(x) = log_a x, 0 < a < 1 is a decreasing function for all x > 0.

Solution:

We have,

=> f(x) = log_a x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(log_a x)$

f'(x) = 1/xloga

As we are given 0 < a < 1,

=> log a < 0

And for x > 0, 1/x > 0

Therefore, f'(x) is,

=> 1/xloga < 0

=> f'(x) < 0

Thus, f(x) is a decreasing function for all x > 0.

Hence proved.

Question 7. Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π).

Solution:

We have,

f(x) = sin x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\sin x)$

f'(x) = cos x

Now for 0 < x < π/2,

=> cos x > 0

=> f'(x) > 0

And for π/2 < x < π,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).

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

Hence proved.

Question 8. Show that f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

Solution:

We have,

f(x) = log sin x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\log\sin x)$

f'(x) = (1/sinx)cosx

f'(x) = cot x

Now for 0 < x < π/2,

=> cot x > 0

=> f'(x) > 0

And for π/2 < x < π,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).

Hence proved.

Question 9. Show that f(x) = x – sin x is increasing for all x ∈ R.

Solution:

We have,

f(x) = x – sin x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x - sin x)$

f'(x) = 1 – cos x

Now, we are given x ∈ R, we get

=> –1 < cos x < 1

=> –1 > cos x > 0

=> f’(x) > 0

Thus, f(x) is increasing on interval x ∈ R.

Hence proved.

Question 10. Show that f(x) = x³ – 15x² + 75x – 50 is an increasing function for all x ∈ R.

Solution:

We have,

f(x) = x³ – 15x² + 75x – 50

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^3 - 15x^2 + 75x - 50)$

f'(x) = 3x² – 30x + 75 – 0

f'(x) = 3x² – 30x + 75

f’(x) = 3(x² – 10x + 25)

f’(x) = 3(x – 5)²

Now, as we are given x ϵ R, we get

=> (x – 5)² > 0

=> 3(x – 5)² > 0

=> f’(x) > 0

Thus, f(x) is increasing on interval x ∈ R.

Hence proved.

Suggest improvement

Class 12 RD Sharma Solutions - Chapter 17 Increasing and Decreasing Functions - Exercise 17.1

Class 12 RD Sharma Solutions - Chapter 17 Increasing and Decreasing Functions - Exercise 17.2 | Set 2

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Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.2 | Set 1

Question 1. Find the intervals in which the following functions are increasing or decreasing.

(i) f(x) = 10 – 6x – 2x2

(ii) f(x) = x2 + 2x – 5

(iii) f(x) = 6 – 9x – x2

(iv) f(x) = 2x3 – 12x2 + 18x + 15

(v) f(x) = 5 + 36x + 3x2 – 2x3

(vi) f(x) = 8 + 36x + 3x2 – 2x3

(vii) f(x) = 5x3 – 15x2 – 120x + 3

(viii) f(x) = x3 – 6x2 – 36x + 2

(ix) f(x) = 2x3 – 15x2 + 36x + 1

(x) f(x) = 2x3 + 9x2 + 12x + 1

(xi) f(x) = 2x3 – 9x2 + 12x – 5

(xii) f(x) = 6 + 12x + 3x2 – 2x3

(i) f(x) = 10 – 6x – 2x²

(ii) f(x) = x² + 2x – 5

(iii) f(x) = 6 – 9x – x²

(iv) f(x) = 2x³ – 12x² + 18x + 15

(v) f(x) = 5 + 36x + 3x² – 2x³

(vi) f(x) = 8 + 36x + 3x² – 2x³

(vii) f(x) = 5x³ – 15x² – 120x + 3

(viii) f(x) = x³ – 6x²– 36x + 2

(ix) f(x) = 2x³ – 15x² + 36x + 1

(x) f(x) = 2x³ + 9x² + 12x + 1

(xi) f(x) = 2x³ – 9x² + 12x – 5

(xii) f(x) = 6 + 12x + 3x² – 2x³

(xiii) f(x) = 2x³ – 24x + 107

(xiv) f(x) = –2x³ – 9x² – 12x + 1

(xv) f(x) = (x – 1) (x – 2)²

(xvi) f(x) = x³ – 12x² + 36x + 17

(xvii) f(x) = 2x³ – 24x + 7

(xviii) f(x) = 3x⁴/10 – 4x³/5 -3x² + 36x/5 + 11

(xix) f(x) = x⁴ – 4x

(xx) f(x) = x⁴/4 + 2/3x³ – 5/2x² – 6x + 7

(xxi) f(x) = x⁴– 4x³ + 4x² + 15

(xxii) f(x) = $5x^{\frac{3}{2}}-3x^\frac{5}{2}$ , x > 0

(xxiii) f(x) = x⁸ + 6x²

(xxiv) f(x) = x³ – 6x² + 9x + 15

(xxv) f(x) = [x(x – 2)]²

(xxvi) f(x) = 3x⁴ – 4x³ – 12x² + 5

(xxvii) f(x) = 3x⁴/2 – 4x³ – 45x² + 51

(xxvii) f(x) = $\log(2+x)-\frac{2x}{2+x}$

Question 2. Determine the values of x for which the function f(x) = x² – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x² – 6x + 9 where the normal is parallel to the line y = x + 5.

Question 4. Show that f(x) = e^2x is increasing on R.

Question 5. Show that f(x) = e^1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Question 6. Show that f(x) = log_a x, 0 < a < 1 is a decreasing function for all x > 0.

Question 10. Show that f(x) = x³ – 15x² + 75x – 50 is an increasing function for all x ∈ R.