RD Sharma Class 12 Ex 17.2 Solutions Chapter 17 Increasing and Decreasing Functions  

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter17
Exercise17.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 17.2 Solutions Chapter 17 Increasing and Decreasing Functions  

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 – 12x+ 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) = 15x– 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 – 6x– 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) = 6x– 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) = 6x– 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) = –6x– 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) = x– 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

=> x1/2(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 – 12x+ 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) = e2x is increasing on R.

Solution:

We have,

=> f(x) = e2x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(e^{2x})

f'(x) = 2e2x

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

=> f’(x) > 0

=> 2e2x > 0

=> e2x > 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) = e1/x, x ≠ 0 is a decreasing function for all x ≠ 0.

Solution:

We have,

=> f(x) = e1/x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(e^{\frac{1}{x}})

f'(x) = -ex/x2 

As x ∈ R, we have,

=> ex > 0

Also, we get,

=> 1/x2 > 0

This means, ex/x2 > 0

=> -ex/x2 < 0

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

Hence proved.

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

Solution:

We have,

=> f(x) = loga 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.

Question 11. Show that f(x) = cos2 x is a decreasing function on (0, π/2).
Solution:

We have,
f(x) = cos2 x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(cos^2 x)
f'(x) = 2 cos x (– sin x)
f'(x) = – sin 2x
Now for 0 < x < π/2,
=> sin 2x > 0
=> – sin 2x < 0
=> f'(x) < 0
Thus, f(x) is decreasing on x ∈ (0, π/2).
Hence proved.
Question 12. Show that f(x) = sin x is an increasing function on (–π/2, π/2).
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 –π/2 < x < π/2,
=> cos x > 0
=> f'(x) > 0
Thus, f(x) is increasing on x ∈ (–π/2, π/2).
Hence proved.
Question 13. Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).
Solution:
We have,
f(x) = cos x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(cos x)
f'(x) = – sin x
Now for 0 < x < π,
=> sin x > 0
=> – sin x < 0
=> f’(x) < 0
And for –π < x < 0,
=> sin x < 0
=> – sin x > 0
=> f’(x) > 0
Therefore, f(x) is decreasing in (0, π) and increasing in (–π, 0).
Hence f(x) is neither increasing nor decreasing in (–π, π).
 Hence proved.
Question 14. Show that f(x) = tan x is an increasing function on (–π/2, π/2).
Solution:
We have,
f(x) = tan x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(tan x)
f'(x) = sec2 x
Now for –π/2 < x < π/2,
=> secx > 0
=> f’(x) > 0
Thus, f(x) is increasing on interval (–π/2, π/2).
Hence proved.
Question 15. Show that f(x) = tan–1 (sin x + cos x) is a decreasing function on the interval (π/4, π /2).
Solution:
We have,
f(x) = tan–1 (sin x + cos x)
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(tan^{-1} (sin x + cos x))
f'(x) = \frac{1}{1+(sinx+cosx)^2}(cosx-sinx)
f'(x) = \frac{cosx-sinx}{1+(sinx+cosx)^2}
f'(x) = \frac{cosx-sinx}{1+sin^2x+cos^2+2sinxcosx}
f'(x) = \frac{cosx-sinx}{1+1+2sinxcosx}
f'(x) = \frac{cosx-sinx}{2+2sinxcosx}
f'(x) = \frac{cosx-sinx}{2(1+sinxcosx)}
Now for π/4 < x < π/2,
=> \frac{cosx-sinx}{2(1+sinxcosx)}      < 0
=> f’(x) < 0
Thus, f(x) is decreasing on interval (π/4, π/2).
Hence proved.
Question 16. Show that the function f(x) = sin (2x + π/4) is decreasing on (3π/8, 5π/8).
Solution:
We have,
f(x) = sin (2x + π/4)
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(sin (2x + π/4))
f'(x) = 2 cos (2x + π/4) 
Now we have, 3π/8 < x < 5π/8
=> 3π/4 < 2x < 5π/4
=> 3π/4  + π/4 < 2x + π/4  < 5π/4 + π/4
=> π < 2x + π/4 + 3π/2
As, 2x + π/4 lies in 3rd quadrant, we get,
=> cos (2x + π/4) < 0
=> 2 cos (2x + π/4) < 0
=> f'(x) < 0
Thus, f(x) is decreasing on interval (3π/8, 5π/8).
Hence proved.
Question 17. Show that the function f(x) = cot–1 (sin x + cos x) is increasing on (0, π/4) and decreasing on (π/4, π/2).
Solution:
We have,
f(x) = cot–1 (sin x + cos x)
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(cot^{-1} (sin x + cos x))
f'(x) = \frac{1}{1+(sinx+cosx)^2}(cosx-sinx)
f'(x) = \frac{cosx-sinx}{1+(sinx+cosx)^2}
f'(x) = \frac{cosx-sinx}{1+sin^2x+cos^2+2sinxcosx}
f'(x) = \frac{cosx-sinx}{1+1+2sinxcosx}
f'(x) = \frac{cosx-sinx}{2+2sinxcosx}
f'(x) = \frac{cosx-sinx}{2(1+sinxcosx)}
Now for π/4 < x < π/2,
=> \frac{cosx-sinx}{2(1+sinxcosx)}      < 0
=> cos x – sin x < 0
=> f’(x) < 0
Also for 0 < x < π/4,
=> \frac{cosx-sinx}{2(1+sinxcosx)}      > 0
=> cos x – sin x > 0
=> f'(x) > 0
Thus, f(x) is increasing on interval (0, π/4) and decreasing on intervals (π/4, π/2).
Hence proved.
Question 18. Show that f(x) = (x – 1) ex + 1 is an increasing function for all x > 0.
Solution:
We have,
f(x) = (x – 1) ex + 1
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}((x - 1) e^x + 1)
f'(x) = ex + (x – 1) ex
f'(x) = ex(1+ x – 1)
f'(x) = x ex
Now for x > 0,
⇒ ex > 0
⇒ x ex > 0
⇒ f’(x) > 0
Thus f(x) is increasing on interval x > 0.
Hence proved.
Question 19. Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).
Solution:
We have,
f(x) = x2 – x + 1
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(x^2 - x + 1)
f'(x) = 2x – 1 + 0
f'(x) = 2x – 1
Now for 0 < x < 1/2, we have
=> 2x – 1 < 0
=> f(x) < 0
Also for 1/2 < x < 1,
=> 2x – 1 > 0
=> f(x) > 0
Thus f(x) is increasing on interval (1/2, 1) and decreasing on interval (0, 1/2).
Hence, the function is neither increasing nor decreasing on (0, 1).
Hence proved.
Question 20. Show that f(x) = x9 + 4x7 + 11 is an increasing function for all x ∈ R.
Solution:
We have,
f(x) = x9 + 4x7 + 11
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(x^9 + 4x^7 + 11)
f'(x) = 9x8 + 28x6 + 0
f'(x) = 9x8 + 28x6 
f'(x) = x6 (9x2 + 28) 
As it is given, x ∈ R, we get,
=> x6 > 0
Also, we can conclude that,
=> 9x2 + 28 > 0
This gives us, f'(x) > 0.
Hence, the function is increasing on the interval x ∈ R.
Hence proved.
Question 21. Show that f(x) = x3 – 6x2 + 12x – 18 is increasing on R.
Solution:
We have,
f(x) = x3 – 6x2 + 12x – 18
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 18)
f'(x) = 3x2 – 12x + 12 –  0
f'(x) = 3x2 – 12x + 12 
f'(x) = 3 (x2 – 4x + 4)
f'(x) = 3 (x – 2)2
Now for x ∈ R, we get,
=> (x – 2)2 > 0
=> 3 (x – 2)2 > 0
=> f'(x) > 0
Hence, the function is increasing on the interval x ∈ R.
Hence proved.
Question 22. State when a function f(x) is said to be increasing on an interval [a, b]. Test whether the function f(x) = x2 – 6x + 3 is increasing on the interval [4, 6]. 
Solution:
A function f(x) is said to be increasing on an interval [a, b] if f(x) > 0.
We have,
f(x) = x2 – 6x + 3
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(x^2 - 6x + 3)
f'(x) = 2x – 6 + 0
f'(x) = 2x – 6
f'(x) = 2(x – 3)
Now for x ∈ [4, 6], we get,
=> 4 ≤ x ≤ 6
=> 1 ≤ (x – 3) ≤ 3
=> x – 3 > 0
=> f'(x) > 0
Thus, the function is increasing on the interval [4, 6].
Hence proved.
Question 23. Show that f(x) = sin x – cos x is an increasing function on (–π/4, π/4).
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
f'(x) = \sqrt{2}[\frac{1}{\sqrt{2}}cosx+\frac{1}{\sqrt{2}}sinx]
f'(x) = \sqrt{2}[sin\frac{\pi}{4}cosx+cos\frac{\pi}{4}sinx]
f'(x) = \sqrt{2}sin(\frac{\pi}{4}+x)
Now we have, x ∈ (–π/4, π/4)
=> –π/4 < x < π/4
=> 0 < (x + π/4) < π/2
=> sin 0 < sin (x + π/4) < sin π/2
=> 0 < sin (x + π/4) < 1
=> sin (x + π/4) > 0
=> √2 sin (x + π/4) > 0
=> f'(x) > 0
Thus, the function is increasing on the interval (–π/4, π/4).
Hence proved.
Question 24. Show that f(x) = tan–1 x – x is a decreasing function on R.
Solution:
We have,
f(x) = tan–1 x – x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(tan^{-1} x - x)
f'(x) = \frac{1}{1+x^2}-1
f'(x) = \frac{1-1-x^2}{1+x^2}
f'(x) = \frac{-x^2}{1+x^2}
Now for x ∈ R, we have,
=> x2 > 0 and 1 + x2 > 0
=> \frac{x^2}{1+x^2}     > 0
=> \frac{-x^2}{1+x^2}     < 0
=> f'(x) < 0
Thus, f(x) is a decreasing function on the interval x ∈ R.
Hence proved.
Question 25. Determine whether f(x) = –x/2 + sin x is increasing or decreasing function on (–π/3, π/3).
Solution:
We have,
f(x) = –x/2 + sin x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(\frac{-x}{2} + sin x)
f'(x) = \frac{-1}{2} + cosx
Now, we have
=> x ∈ (–π/3, π/3)
=> –π/3 < x < π/3
=> cos (–π/3) < cos x < cos (π/3)
=> 1/2 < cos x < 1/2
=> \frac{-1}{2} + cosx     > 0
=> f'(x) > 0
Thus, f(x) is a increasing function on the interval x ∈ (–π/3, π/3).
Hence proved.

Question 26. Find the intervals in which f(x) = log (1 + x) – x/(1 + x) is increasing or decreasing.

Solution:

We have,

f(x) = log (1 + x) – x/(1 + x)

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(\log(1+x) -\frac{x}{(1+x)})

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

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

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

f'(x) = \frac{1+x-1}{(1+x)^2}

f'(x) = \frac{x}{(1+x)^2}

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

=> f'(x) = 0

=> \frac{x}{(1+x)^2}    = 0

=> x = 0

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

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

Thus f(x) is increasing in (0, ∞) and decreasing in (–∞, –1) ∪ (–1, 0).

Question 27. Find the intervals in which f(x) = (x + 2)e–x is increasing or decreasing.

Solution:

We have,

f(x) = (x + 2)e–x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}((x + 2)e^{-x})

f'(x) = e–x – e–x (x + 2)

f'(x) = e–x (1 – x – 2)

f'(x) = e–x (x + 1)

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

=> f'(x) = 0

=> e–x (x + 1) = 0

=> x = –1

Clearly, f'(x) > 0 if x < –1.

Also, f'(x) < 0 if x > –1.

Thus f(x) is increasing in (–∞, –1) and decreasing in (–1, ∞).

Question 28. Show that the function f given by f(x) = 10x is increasing for all x.

Solution:

We have,

f(x) = 10x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(10^x)

f'(x) = 10x log 10

Now we have, x ∈ R, we get

=> 10x > 0

=> 10x log 10 > 0

=> f'(x) > 0

Thus, f(x) is increasing for all x.

Hence proved.

Question 29. Prove that the function f given by f(x) = x – [x] is increasing in (0, 1).

Solution:

We have,

f(x) = x – [x]

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(x - [x])

f'(x) = 1

Now we have,

=> 1 > 0

=> f'(x) > 0

Thus, f(x) is increasing in the interval (0, 1).

Hence proved.

Question 30. Prove that the function f(x) = 3x5 + 40x3 + 240x is increasing on R.

Solution:

We have,

f(x) = 3x5 + 40x3 + 240x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(3x^5 + 40x^3 + 240x)

f'(x) = 15x4 + 120x2 + 240

f'(x) = 15 (x4 + 8x2 + 16)

f'(x) = 15 (x2 + 4)2

Now we know,

=> (x2 + 4)2 > 0

=> 15 (x2 + 4)2 > 0

=> f'(x) > 0

Thus, the given f(x) is increasing on R.

Hence proved.

Question 31. Prove that the function f given by f(x) = log cos x is strictly increasing on (–π/2, 0) and strictly decreasing on (0, π/2).

Solution:

We have,

f(x) = log cos x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(\log \cos x)

f'(x) = \frac{1}{cosx}(-sinx)

f'(x) = \frac{-sinx}{cosx}

f'(x) = – tan x

Now for x ∈ (0, π/2), we get

=> 0 < x < π/2

=> tan 0 < tan x < tan π/2

=> 0 < tan x < 1

=> tan x > 0

=> – tan x < 0

=> f'(x) < 0

Also for x ∈ (–π/2, 0), we have,

=> –π/2 < x < 0

=> tan (–π/2) < tan x <  tan 0

=> –1 < tan x < 0

=> tan x < 0

=> – tan x > 0

=> f'(x) > 0

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

Hence proved.

Question 32. Show that the function f given by f(x) = x3 – 3x2 + 4x is strictly increasing on R.

Solution:

We have,

f(x) = x3 – 3x2 + 4x 

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 4x)

f'(x) = 3x2 – 6x + 4

f'(x) = 3 (x2 – 2x + 1) + 1

f'(x) = 3 (x – 1)2 + 1

Now, we know,

=> (x – 1)> 0

=> 3 (x – 1)2 > 0

=> 3 (x – 1)2 + 1 > 0

=> f'(x) > 0

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

Hence proved.

Question 33. Show that the function f given by f(x) = cos x is strictly decreasing in (0, π), increasing in (π, 2π) and neither increasing or decreasing in (0, 2π).  

Solution:

We have,

f(x) = cos x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(cos x)

f'(x) = – sin x

Now for x ∈ (0, π), we get

=> 0 < x < π

=> sin 0 < sin x < sin π

=> 0 < sin x < 0

=> sin x > 0

=> – sin x < 0

=> f'(x) < 0

Also for x ∈ (π, 2π), we get

=> π < x < 2π

=> sin 0 < sin x < sin π

=> 0 < sin x < 0

=> sin x < 0

=> – sin x > 0

=> f'(x) > 0

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

So, the function is neither increasing or decreasing in (0, 2π). 

Hence proved.

Question 34. Show that f(x) = x2 – x sin x is an increasing function on (0, π/2).  

Solution:

We have,

f(x) = x2 – x sin x 

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(x^2 - x sin x)

f'(x) = 2x – (x cos x + sin x) 

f'(x) = 2x – x cos x – sin x

Now for x ∈ (0, π/2), we have

=> 0 ≤ sin x ≤ 1

=> 0 ≤ cos x ≤ 1

So, this implies,

=> 2x – x cos x – sin x > 0

=> f'(x) > 0

Thus, f(x) is an increasing function on the interval (0, π/2).

Hence proved.

Question 35. Find the value(s) of a for which f(x) = x3 – ax is an increasing function on R. 

Solution:

We have,

f(x) = x3 – ax

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(x^3 - ax)

f'(x) = 3x2 – a

Now we are given that f(x) = x3 – ax is an increasing function on R, we get

=> f'(x) > 0

=> 3x2 – a > 0

=> a < 3x2

The critical point for 3x2 = 0 will be 0.

So, we get a ≤ 0.

Therefore, the values of a must be less than or equal to 0.

Question 36. Find the value of b for which the function f(x) = sin x – bx + c is a decreasing function on R.

Solution:

We have,

f(x) = sin x – bx + c

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(sin x - bx + c)

f'(x) = cos x – b + 0

f'(x) = cos x – b 

Now we are given that f(x) = sin x – bx + c is a decreasing function on R, we get

=> f'(x) < 0

=> cos x – b < 0

=> b > cos x

The critical point for cos x = 0 will be 1.

So, we get b ≥ 1.

Therefore, the values of b must be greater than or equal to 1.

Question 37. Show that f(x) = x + cos x – a is an increasing function on R for all values of a.

Solution:

We have,

f(x) = x + cos x – a

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(x + cos x - a)

f'(x) = 1 – sin x

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

f'(x) = (sin\frac{x}{2}-cos\frac{x}{2})^2

Now for x ∈ R, we have

=> (sin\frac{x}{2}-cos\frac{x}{2})^2   > 0

=> f'(x) > 0

Thus, the f(x) is an increasing function on R for all values of a.

Hence proved.

Question 38. Let f defined on [0, 1] be twice differentiable such that |f”(x)| ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that f'(x) < 1 for all x ∈ [0, 1].

Solution:

As f(0) = f(1) and f is differentiable, we can apply Rolle’s theorem here. So, we get

f(c) = 0 for some c ∈ [0, 1].

On applying Lagrange’s mean value theorem, we get,

For point c and x ∈ [0, 1], so we have

=> \frac{|f'(x)-f'(c)|}{x-c}=f''(d)

=> \frac{|f'(x)-0|}{x-c}=f''(d)

=> \frac{|f'(x)|}{x-c}=f''(d)

As we are given that |f”(d)| ≤ 1 for x ∈ [0, 1], we get

=> \frac{|f'(x)|}{x-c}   ≤ 1

=> |f'(x)| ≤ x – c

Now as both x and c lie in [0, 1], therefore x  – c ∈ (0, 1).

This gives us, |f'(x)| < 1 for all x ∈ (0, 1).

Hence proved.

Question 39. Find the interval in which f(x) is increasing or decreasing:

(i) f(x) = x |x|, x ∈ R

Solution:

We have,

f(x) = x |x|, x ∈ R

=> f(x)=\begin{cases}-x^2,x<0\\x^2,x>0\end{cases}” height=”79″ width=”229″></p><p>=> <img src=(ii) f(x) = sin x + |sin x|, 0 < x ≤ 2π

Solution:

We have,

f(x) = sin x + |sin x|, 0 < x ≤ 2π

=> f(x)=\begin{cases}2sinx,0 < x < π\\0,\pi < x \le 2π\end{cases}

=> f'(x)=\begin{cases}2cosx,0 < x < π\\0,\pi < x \le 2π\end{cases}

The function cos x is positive between the interval (0, π/2).

Therefore, the function is increasing in the interval (0, π/2).

Also, the function cos x is negative between the interval (π/2, π).

Therefore, the function is decreasing in the interval (0, π/2).

Now for π ≤ x ≤ 2π, value of f'(x) is 0.

Hence the function is neither increasing nor decreasing in the interval (π, 2π).

(iii) f(x) = sin x (1 + cos x), 0 < x ≤ π/2

Solution:

We have,

f(x) = sin x (1 + cos x)

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(sin x (1 + cos x))

f'(x) = sinx(-sinx)+(1+cosx)(cosx)

f'(x) = –sin2 x + cos x + cos2

f'(x) = cosx – sin2 x + cos x

f'(x) = cos2 x – (1 – cos2 x) + cos x

f'(x) = cos2 x – 1 + cos2 x + cos x

f'(x) = 2 cos2 x + cos x – 1

f'(x) = 2 cos2 x + 2 cos x – cos x – 1

f'(x) = 2 cos x (cos x + 1) – 1 (cos x + 1)

f'(x) = (2 cos x – 1) (cos x + 1)

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

=> f'(x) > 0

=> (2 cos x – 1) (cos x + 1) > 0

=> 0 < x < π/3

=> x ∈ (0, π/3)

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

=> f'(x) < 0

=> (2 cos x – 1) (cos x + 1) > 0

=> π/3 < x < π/2

=> x ∈ (π/3, π/2)

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

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