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

• Last Updated : 02 Jun, 2021

### (i) f(x) = 10 â€“ 6x â€“ 2x2

Solution:

We are given,

f(x) = 10 â€“ 6x â€“ 2x2

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 â€“ 6 â€“ 4x

f'(x) = â€“ 6 â€“ 4x

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

=> f'(x) > 0

=> â€“ 6 â€“ 4x > 0

=> â€“ 4x > 6

=> x < â€“6/4

=> x < â€“3/2

=> x âˆˆ (â€“âˆž, â€“3/2)

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

=> f'(x) < 0

=> â€“ 6 â€“ 4x < 0

=> â€“ 4x < 6

=> x > â€“6/4

=> x > â€“3/2

=> x âˆˆ ( â€“3/2, âˆž)

Thus, f(x) is increasing on the interval x âˆˆ (â€“âˆž, â€“3/2) and decreasing on the interval x âˆˆ ( â€“3/2, âˆž).

### (ii) f(x) = x2 + 2x â€“ 5

Solution:

We are given,

f(x) = x2 + 2x â€“ 5

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2x + 2 â€“ 0

f'(x) = 2x + 2

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

=> f'(x) > 0

=> 2x + 2 > 0

=> 2x > â€“2

=> x > â€“2/2

=> x > â€“1

=> x âˆˆ (â€“1, âˆž)

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

=> f'(x) < 0

=> 2x + 2 < 0

=> 2x < â€“2

=> x < â€“2/2

=> x < â€“1

=> x âˆˆ (â€“âˆž, â€“1)

Thus, f(x) is increasing on the interval x âˆˆ (â€“1, âˆž) and decreasing on the interval x âˆˆ ( â€“âˆž, â€“1).

### (iii) f(x) = 6 â€“ 9x â€“ x2

Solution:

We are given,

f(x) = 6 â€“ 9x â€“ x2

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 â€“ 9 â€“ 2x

f'(x) = â€“ 9 â€“ 2x

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

=> f'(x) > 0

=> â€“9 â€“ 2x > 0

=> â€“2x > 9

=> x > â€“9/2

=> x âˆˆ (â€“9/2, âˆž)

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

=> f'(x) < 0

=> â€“9 â€“ 2x < 0

=> â€“2x < 9

=> x < â€“9/2

=> x âˆˆ (â€“âˆž, â€“9/2)

Thus, f(x) is increasing on the interval x âˆˆ (â€“9/2, âˆž) and decreasing on the interval x âˆˆ ( â€“âˆž, â€“9/2).

### (iv) f(x) = 2x3 â€“ 12x2 + 18x + 15

Solution:

We are given,

f(x) = 2x3 â€“ 12x2 + 18x + 15

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x2 â€“ 24x + 18 + 0

f'(x) = 6x2 â€“ 24x + 18

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

=> f'(x) = 0

=> 6x2 â€“ 24x + 18 = 0

=> 6 (x2 â€“ 4x + 3) = 0

=> x2 â€“ 4x + 3 = 0

=> x2 â€“ 3x â€“ x + 3 = 0

=> x (x â€“ 3) â€“ 1 (x â€“ 3) = 0

=> (x â€“ 1) (x â€“ 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 âˆˆ (â€“âˆž, 1)âˆª (3, âˆž) and decreasing on the interval x âˆˆ (1, 3).

### (v) f(x) = 5 + 36x + 3x2 â€“ 2x3

Solution:

We are given,

f(x) = 5 + 36x + 3x2 â€“ 2x3

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 36 + 6x â€“ 6x2

f'(x) = 36 + 6x â€“ 6x2

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

=> f'(x) = 0

=> 36 + 6x â€“ 6x2 = 0

=> 6 (â€“ x2 + x + 6) = 0

=> 6 (â€“x2 + 3x â€“ 2x + 6) = 0

=> â€“x2 + 3x â€“ 2x + 6 = 0

=> x2 â€“ 3x + 2x â€“ 6 = 0

=> (x â€“ 3) (x + 2) = 0

=> x = 3, â€“ 2

Clearly, fâ€™(x) > 0 if â€“2 < x < 3.

Also fâ€™(x) < 0 if x < â€“2 and x > 3.

Thus, f(x) is increasing on x âˆˆ (â€“2, 3) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“2) âˆª (3, âˆž).

### (vi) f(x) = 8 + 36x + 3x2 â€“ 2x3

Solution:

We are given,

f(x) = 8 + 36x + 3x2 â€“ 2x3

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 36 + 6x â€“ 6x2

f'(x) = 36 + 6x â€“ 6x2

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

=> f'(x) = 0

=> 36 + 6x â€“ 6x2 = 0

=> 6 (â€“ x2 + x + 6) = 0

=> 6 (â€“x2 + 3x â€“ 2x + 6) = 0

=> â€“x2 + 3x â€“ 2x + 6 = 0

=> x2 â€“ 3x + 2x â€“ 6 = 0

=> (x â€“ 3) (x + 2) = 0

=> x = 3, â€“2

Clearly, fâ€™(x) > 0 if â€“2 < x < 3.

Also fâ€™(x) < 0 if x < â€“2 and x > 3.

Thus, f(x) is increasing on x âˆˆ (â€“2, 3) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“2) âˆª (3, âˆž).

### (vii) f(x) = 5x3 â€“ 15x2 â€“ 120x + 3

Solution:

We are given,

f(x) = 5x3 â€“ 15x2 â€“ 120x + 3

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 15x2 â€“ 30x â€“ 120 + 0

f'(x) = 15x2 â€“ 30x â€“ 120

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

=> f'(x) = 0

=> 15x2 â€“ 30x â€“ 120 = 0

=> 15(x2 â€“ 2x â€“ 8) = 0

=> 15(x2 â€“ 4x + 2x â€“ 8) = 0

=> x2 â€“ 4x + 2x â€“ 8 = 0

=> (x â€“ 4) (x + 2) = 0

=> x = 4, â€“2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 4.

Also fâ€™(x) < 0 if â€“2 < x < 4.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž,â€“2) âˆª (4, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“2, 4).

### (viii) f(x) = x3 â€“ 6x2 â€“ 36x + 2

Solution:

We are given,

f(x) = x3 â€“ 6x2 â€“ 36x + 2

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x2 â€“ 12x â€“ 36 + 0

f'(x) = 3x2 â€“ 12x â€“ 36

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

=> f'(x) = 0

=> 3x2 â€“ 12x â€“ 36 = 0

=> 3(x2 â€“ 4x â€“ 12) = 0

=> 3(x2 â€“ 6x + 2x â€“ 12) = 0

=> x2 â€“ 6x + 2x â€“ 12 = 0

=> (x â€“ 6) (x + 2) = 0

=> x = 6, â€“2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 6.

Also fâ€™(x) < 0 if â€“2< x < 6

Thus, f(x) is increasing on x âˆˆ (â€“âˆž,â€“2) âˆª (6, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“2, 6).

### (ix) f(x) = 2x3 â€“ 15x2 + 36x + 1

Solution:

We are given,

f(x) = 2x3 â€“ 15x2 + 36x + 1

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x2 â€“ 30x + 36 + 0

f'(x) = 6x2 â€“ 30x + 36

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

=> f'(x) = 0

=> 6x2 â€“ 30x + 36 = 0

=> 6 (x2 â€“ 5x + 6) = 0

=> 6(x2 â€“ 3x â€“ 2x + 6) = 0

=> x2 â€“ 3x â€“ 2x + 6 = 0

=> (x â€“ 3) (x â€“ 2) = 0

=> x = 3, 2

Clearly, fâ€™(x) > 0 if x < 2 and x > 3.

Also fâ€™(x) < 0 if 2 < x < 3.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, 2) âˆª (3, âˆž) and f(x) is decreasing on interval x âˆˆ (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 = â€“1, â€“2

Clearly, fâ€™(x) > 0 if â€“2 < x < â€“1.

Also fâ€™(x) < 0 if x < â€“1 and x > â€“2.

Thus, f(x) is increasing on x âˆˆ (â€“2,â€“1) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“2) âˆª (â€“2, âˆž).

### (xi) f(x) = 2x3 â€“ 9x2 + 12x â€“ 5

Solution:

We are given,

f(x) = 2x3 â€“ 9x2 + 12x â€“ 5

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x2 â€“ 18x + 12 â€“ 0

f'(x) = 6x2 â€“ 18x + 12

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

=> f'(x) = 0

=> 6x2 â€“ 18x + 12 = 0

=> 6 (x2 â€“ 3x + 2) = 0

=> 6(x2 â€“ 2x â€“ x + 2) = 0

=> x2 â€“ 2x â€“ x + 2 = 0

=> (x â€“ 2) (x â€“ 1) = 0

=> x = 1, 2

Clearly, fâ€™(x) > 0 if x < 1 and x > 2.

Also fâ€™(x) < 0 if 1 < x < 2.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, 1) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (1, 2).

### (xii) f(x) = 6 + 12x + 3x2 â€“ 2x3

Solution:

We are given,

f(x) = 6 + 12x + 3x2 â€“ 2x3

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 0 + 12 + 6x â€“ 6x2

f'(x) = 12 + 6x â€“ 6x2

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

=> f'(x) = 0

=> 12 + 6x â€“ 6x2 = 0

=> 6 (â€“x2 + x + 2) = 0

=> x2 â€“ x â€“ 2 = 0

=> x2 â€“ 2x + x â€“ 2 = 0

=> (x â€“ 2) (x + 1) = 0

=> x = 2, â€“1

Clearly, fâ€™(x) > 0 if â€“1 < x < 2.

Also fâ€™(x) < 0 if x < â€“1 and x > 2.

Thus, f(x) is increasing on x âˆˆ (â€“1, 2) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“1) âˆª (2, âˆž).

### (xiii) f(x) = 2x3 â€“ 24x + 107

Solution:

We are given,

f(x) = 2x3 â€“ 24x + 107

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x2 â€“ 24 + 0

f'(x) = 6x2 â€“ 24

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

=> f'(x) = 0

=> 6x2 â€“ 24 = 0

=> 6x2 = 24

=> x2 = 4

=> x = 2, â€“2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 2.

Also fâ€™(x) < 0 if â€“2 < x < 2.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, â€“2) âˆª (2, âˆž), and f(x) is decreasing on interval x âˆˆ (â€“2, 2).

### (xiv) f(x) = â€“2x3 â€“ 9x2 â€“ 12x + 1

Solution:

We are given,

f(x) = â€“2x3 â€“ 9x2 â€“ 12x + 1

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = â€“6x2 â€“ 18x â€“ 12 + 0

f'(x) = â€“6x2 â€“ 18x â€“ 12

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

=> f'(x) = 0

=> â€“6x2 â€“ 18x â€“ 12 = 0

=> 6 (â€“x2 â€“ 3x â€“ 2) = 0

=> x2 + 3x + 2 = 0

=> x2 + 2x + x + 2 = 0

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

=> x = â€“2, â€“1

Clearly, fâ€™(x) > 0 if x < â€“1 and x > â€“2.

Also, fâ€™(x) < 0 if â€“2 < x < â€“1.

Thus, f(x) is increasing on x âˆˆ (â€“2, â€“1) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“2) âˆª (â€“1, âˆž).

### (xv) f(x) = (x â€“ 1) (x â€“ 2)2

Solution:

We are given,

f(x) = (x â€“ 1) (x â€“ 2)2

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = (x â€“ 2)2 + 2 (x â€“ 1) (x â€“ 2)

f'(x) = (x â€“ 2) (x â€“ 2 + 2x â€“ 2)

f'(x) = (x â€“ 2) (3x â€“ 4)

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

=> f'(x) = 0

=> (x â€“ 2) (3x â€“ 4) = 0

=> x = 2, 4/3

Clearly, fâ€™(x) > 0 if x < 4/3 and x > 2.

Also, fâ€™(x) < 0 if 4/3 < x < 2.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, 4/3) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (4/3, 2).

### (xvi) f(x) = x3 â€“ 12x2 + 36x + 17

Solution:

We are given,

f(x) = x3 â€“ 12x2 + 36x + 17

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x2 â€“ 24x + 36 + 0

f'(x) = 3x2 â€“ 24x + 36

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

=> f'(x) = 0

=> 3x2 â€“ 24x + 36 = 0

=> 3 (x2 â€“ 8x + 12) = 0

=> x2 â€“ 8x + 12 = 0

=> x2 â€“ 6x â€“ 2x + 12 = 0

=> (x â€“ 6) (x â€“ 2) = 0

=> x = 6, 2

Clearly, fâ€™(x) > 0 if x < 2 and x > 6.

Also, fâ€™(x) < 0 if 2 < x < 6.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, 2) âˆª (6, âˆž) and f(x) is decreasing on interval x âˆˆ (2, 6).

### (xvii) f(x) = 2x3 â€“ 24x + 7

Solution:

We are given,

f(x) = 2x3 â€“ 24x + 7

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 6x2 â€“ 24 + 0

f'(x) = 6x2 â€“ 24

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

=> f'(x) = 0

=> 6x2 â€“ 24 = 0

=> 6x2 = 24

=> x2 = 4

=> x = 2, â€“2

Clearly, fâ€™(x) > 0 if x < â€“2 and x > 2.

Also fâ€™(x) < 0 if â€“2 < x < 2.

Thus, f(x) is increasing on x âˆˆ (â€“âˆž, â€“2) âˆª (2, âˆž), and f(x) is decreasing on interval x âˆˆ (â€“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, â€“2, 3

Clearly, fâ€™(x) > 0 if â€“2 < x < 1 and if x > 3

Also fâ€™(x) < 0 if 1 < x < 3.

Thus, f(x) is increasing on x âˆˆ (3, âˆž) and f(x) is decreasing on interval x âˆˆ (1, 3).

### (xix) f(x) = x4 â€“ 4x

Solution:

We are given,

f(x) = x4 â€“ 4x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 4x3 â€“ 4

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

=> f'(x) = 0

=> 4x3 â€“ 4 = 0

=> 4 (x3 â€“ 1) = 0

=> x3 â€“ 1 = 0

=> x3 = 1

=> x = 1

Clearly, fâ€™(x) > 0 if x > 1.

Also fâ€™(x) < 0 if x < 1.

Thus, f(x) is increasing on x âˆˆ (1, âˆž), and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 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 â€“ 5x â€“ 6

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

=> f'(x) = 0

=> x3 + 2x2 â€“ 5x â€“ 6 = 0

=> (x + 1) (x + 3) (x â€“ 2) = 0

=> x = â€“1, â€“3, 2

Clearly f'(x) > 0 if â€“3 < x < â€“1 and x > 2.

Also f'(x) < 0 if x < â€“3 and â€“1 < x < 2.

Thus, f(x) is increasing on x âˆˆ (â€“3, â€“1) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“3) âˆª (â€“1, 2).

### (xxi) f(x) = x4 â€“ 4x3 + 4x2 + 15

Solution:

We have,

f(x) = x4 â€“ 4x3 + 4x2 + 15

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 4x3 â€“ 12x2 + 8x + 0

f'(x) = 4x3 â€“ 12x2 + 8x

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

=> f'(x) = 0

=> 4x3 â€“ 12x2 + 8x = 0

=> 4x (x2 â€“ 3x + 2) = 0

=> 4x (x â€“ 2) (x â€“ 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 âˆˆ (0, 1) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 0) âˆª (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 âˆˆ (0, 1) and f(x) is decreasing on interval x âˆˆ (1, âˆž).

### (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 âˆˆ (0, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 0).

### (xxiv) f(x) = x3 â€“ 6x2 + 9x + 15

Solution:

We are given,

f(x) = x3 â€“ 6x2 + 9x + 15

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x2 â€“ 12x + 9 + 0

f'(x) = 3x2 â€“ 12x + 9

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

=> f'(x) = 0

=> 3x2 â€“ 12x + 9 = 0

=> 3 (x2 â€“ 4x + 3) = 0

=> x2 â€“ 4x + 3 = 0

=> x2 â€“ 3x â€“ x + 3 = 0

=> (x â€“ 3) (x â€“ 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 âˆˆ (â€“âˆž, 1) âˆª (3, âˆž) and f(x) is decreasing on interval x âˆˆ (1, 3).

### (xxv) f(x) = [x(x â€“ 2)]2

Solution:

We are given,

f(x) = [x(x â€“ 2)]2

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

f'(x) = 2 (x2 â€“ 2x) (2x â€“ 2)

f'(x) = 4x (x â€“ 2) (x â€“ 1)

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

=> f'(x) = 0

=> 4x (x â€“ 2) (x â€“ 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 âˆˆ (0, 1) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 0) âˆª (1, 2).

### (xxvi) f(x) = 3x4 â€“ 4x3 â€“ 12x2 + 5

Solution:

We are given,

f(x) = 3x4 â€“ 4x3 â€“ 12x2 + 5

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 12x3 â€“ 12x2 â€“ 24x

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

=> f'(x) = 0

=> 12x3 â€“ 12x2 â€“ 24x = 0

=> 12x (x2 â€“ x â€“ 2) = 0

=> 12x (x + 1) (x â€“ 2) = 0

=> x = 0, â€“1, 2

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

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

Thus, f(x) is increasing on x âˆˆ (â€“1, 0) âˆª (2, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“1) âˆª (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 â€“ 12x2 â€“ 90x

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

=> f'(x) = 0

=> 6x3 â€“ 12x2 â€“ 90x = 0

=> 6x (x2 â€“ 2x â€“ 15) = 0

=> 6x (x + 3) (x â€“ 5) = 0

=> x = 0, â€“3, 5

Clearly f'(x) > 0 if â€“3 < x < 0 and x > 5.

Also f'(x) < 0 if x < â€“3 and 0< x < 5.

Thus, f(x) is increasing on x âˆˆ (â€“3, 0) âˆª (5, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, â€“3) âˆª (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 âˆˆ (2, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 2).

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

Solution:

Given f(x) = x2 â€“ 6x + 9

On differentiating both sides with respect to x, we get

=> fâ€™(x) = 2x â€“ 6

=> fâ€™(x) = 2(x â€“ 3)

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

=> fâ€™(x) = 0

=> 2(x â€“ 3) = 0

=> (x â€“ 3) = 0

=> x = 3

Clearly, fâ€™(x) > 0 if x > 3.

Also fâ€™(x) < 0 if x < 3.

Thus, f(x) is increasing on (3, âˆž) and f(x) is decreasing on interval x âˆˆ (â€“âˆž, 3).

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

Slope of this curve is given by,

=> m1 = dy/dx

=> m1 = 2x â€“ 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 = â€“1

=> 2x â€“ 6 = â€“1

=> 2x = 5

=> x = 5/2

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

=> y = (5/2)2 â€“ 6 (5/2) + 9

=> y = 25/4 â€“ 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 â€“ cos x, where 0 < x < 2Ï€ is increasing or decreasing.

Solution:

We have,

f(x) = sin x â€“ cos x

On differentiating both sides with respect to x, we get

f'(x) = \frac{d}{dx}(sin x â€“ 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 = â€“1

=> x = 3Ï€/4 , 7Ï€/4

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

Also f'(x) < 0 if 3Ï€/4 < x < 7Ï€/4.

Thus, f(x) is increasing on x âˆˆ (0, 3Ï€/4) âˆª (7Ï€/4, 2Ï€) and f(x) is decreasing on interval x âˆˆ (3Ï€/4, 7Ï€/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

=> fâ€™(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 â‰  0 is a decreasing function for all x â‰  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 âˆˆ 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 â‰  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, Ï€/2) and decreasing on (Ï€/2, Ï€) and neither increasing nor decreasing in (0, Ï€).

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 < Ï€/2,

=> cos x > 0

=> f'(x) > 0

And for Ï€/2 < x < Ï€,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x âˆˆ (0, Ï€/2) and f(x) is decreasing on interval x âˆˆ (Ï€/2, Ï€).

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

Hence proved.

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

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 < Ï€/2,

=> cot x > 0

=> f'(x) > 0

And for Ï€/2 < x < Ï€,

=> cos x < 0

=> f'(x) < 0

Thus, f(x) is increasing on x âˆˆ (0, Ï€/2) and f(x) is decreasing on interval x âˆˆ (Ï€/2, Ï€).

Hence proved.

### Question 9. Show that f(x) = x â€“ sin x is increasing for all x âˆˆ R.

Solution:

We have,

f(x) = x â€“ sin x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 1 â€“ cos x

Now, we are given x âˆˆ R, we get

=> â€“1 < cos x < 1

=> â€“1 > cos x > 0

=> fâ€™(x) > 0

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

Hence proved.

### Question 10. Show that f(x) = x3 â€“ 15x2 + 75x â€“ 50 is an increasing function for all x âˆˆ R.

Solution:

We have,

f(x) = x3 â€“ 15x2 + 75x â€“ 50

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x2 â€“ 30x + 75 â€“ 0

f'(x) = 3x2 â€“ 30x + 75

fâ€™(x) = 3(x2 â€“ 10x + 25)

fâ€™(x) = 3(x â€“ 5)2

Now, as we are given x Ïµ R, we get

=> (x â€“ 5)2 > 0

=> 3(x â€“ 5)2 > 0

=> fâ€™(x) > 0

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

Hence proved.

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