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

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  • Last Updated : 02 Jun, 2021
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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

=> 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 – 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) = 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.


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