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

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

### (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) =

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) =

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) =

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) =

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) =

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) =

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) =

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 – 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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) = , x > 0

Solution:

We have,

f(x) =

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

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

=> f'(x) = 0

=>  = 0

=>  = 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) =

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) =

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) =

f'(x) =

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) =

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) =

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) =

Solution:

We have,

f(x) =

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =

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) =
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) =
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) =
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) =
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) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
Now for π/4 < x < π/2,
=>  < 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) =
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) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
f'(x) =
Now for π/4 < x < π/2,
=>  < 0
=> cos x – sin x < 0
=> f’(x) < 0
Also for 0 < x < π/4,
=>  > 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) =
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) =
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) =
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) =
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) =
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) =
f'(x) = cos x + sin x
f'(x) =
f'(x) =
f'(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) =
f'(x) =
f'(x) =
f'(x) =
Now for x ∈ R, we have,
=> x2 > 0 and 1 + x2 > 0
=>  > 0
=>  < 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) =
f'(x) =
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
=>  > 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) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

f'(x) =

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

=> f'(x) = 0

=>  = 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) =

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) =

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) =

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) =

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) =

f'(x) =

f'(x) =

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) =

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) =

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) =

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) =

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) =

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) =

f'(x) = 1 – sin x

f'(x) =

f'(x) =

Now for x ∈ R, we have

=>  > 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

=>

=>

=>

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

=>  ≤ 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.

### (i) f(x) = x |x|, x ∈ R

Solution:

We have,

f(x) = x |x|, x ∈ R

=> (ii) f(x) = sin x + |sin x|, 0 < x ≤ 2π

Solution:

We have,

f(x) = sin x + |sin x|, 0 < x ≤ 2π

=>

=>

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) =

f'(x) =

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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