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

Last Updated : 15 Feb, 2022

### Question 11. Show that f(x) = cos2 x is a decreasing function on (0, Ï€/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 (â€“ sin x)

f'(x) = â€“ sin 2x

Now for 0 < x < Ï€/2,

=> sin 2x > 0

=> â€“ sin 2x < 0

=> f'(x) < 0

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

Hence proved.

### Question 12. Show that f(x) = sin x is an increasing function on (â€“Ï€/2, Ï€/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 â€“Ï€/2 < x < Ï€/2,

=> cos x > 0

=> f'(x) > 0

Thus, f(x) is increasing on x âˆˆ (â€“Ï€/2, Ï€/2).

Hence proved.

### Question 13. Show that f(x) = cos x is a decreasing function on (0, Ï€), increasing in (â€“Ï€, 0) and neither increasing nor decreasing in (â€“Ï€, Ï€).

Solution:

We have,

f(x) = cos x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = â€“ sin x

Now for 0 < x < Ï€,

=> sin x > 0

=> â€“ sin x < 0

=> fâ€™(x) < 0

And for â€“Ï€ < x < 0,

=> sin x < 0

=> â€“ sin x > 0

=> fâ€™(x) > 0

Therefore, f(x) is decreasing in (0, Ï€) and increasing in (â€“Ï€, 0).

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

Hence proved.

### Question 14. Show that f(x) = tan x is an increasing function on (â€“Ï€/2, Ï€/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 â€“Ï€/2 < x < Ï€/2,

=> sec2 x > 0

=> fâ€™(x) > 0

Thus, f(x) is increasing on interval (â€“Ï€/2, Ï€/2).

Hence proved.

### Question 15. Show that f(x) = tanâ€“1 (sin x + cos x) is a decreasing function on the interval (Ï€/4, Ï€ /2).

Solution:

We have,

f(x) = tanâ€“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 Ï€/4 < x < Ï€/2,

=>  < 0

=> fâ€™(x) < 0

Thus, f(x) is decreasing on interval (Ï€/4, Ï€/2).

Hence proved.

### Question 16. Show that the function f(x) = sin (2x + Ï€/4) is decreasing on (3Ï€/8, 5Ï€/8).

Solution:

We have,

f(x) = sin (2x + Ï€/4)

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2 cos (2x + Ï€/4)

Now we have, 3Ï€/8 < x < 5Ï€/8

=> 3Ï€/4 < 2x < 5Ï€/4

=> 3Ï€/4  + Ï€/4 < 2x + Ï€/4  < 5Ï€/4 + Ï€/4

=> Ï€ < 2x + Ï€/4 + 3Ï€/2

As, 2x + Ï€/4 lies in 3rd quadrant, we get,

=> cos (2x + Ï€/4) < 0

=> 2 cos (2x + Ï€/4) < 0

=> f'(x) < 0

Thus, f(x) is decreasing on interval (3Ï€/8, 5Ï€/8).

Hence proved.

### Question 17. Show that the function f(x) = cotâ€“1 (sin x + cos x) is increasing on (0, Ï€/4) and decreasing on (Ï€/4, Ï€/2).

Solution:

We have,

f(x) = cotâ€“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 Ï€/4 < x < Ï€/2,

=>  < 0

=> cos x â€“ sin x < 0

=> fâ€™(x) < 0

Also for 0 < x < Ï€/4,

=>  > 0

=> cos x â€“ sin x > 0

=> f'(x) > 0

Thus, f(x) is increasing on interval (0, Ï€/4) and decreasing on intervals (Ï€/4, Ï€/2).

Hence proved.

### Question 18. Show that f(x) = (x â€“ 1) ex + 1 is an increasing function for all x > 0.

Solution:

We have,

f(x) = (x â€“ 1) ex + 1

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = ex + (x â€“ 1) ex

f'(x) = ex(1+ x â€“ 1)

f'(x) = x ex

Now for x > 0,

â‡’ ex > 0

â‡’ x ex > 0

â‡’ fâ€™(x) > 0

Thus f(x) is increasing on interval x > 0.

Hence proved.

### Question 19. Show that the function x2 â€“ x + 1 is neither increasing nor decreasing on (0, 1).

Solution:

We have,

f(x) = x2 â€“ x + 1

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2x â€“ 1 + 0

f'(x) = 2x â€“ 1

Now for 0 < x < 1/2, we have

=> 2x â€“ 1 < 0

=> f(x) < 0

Also for 1/2 < x < 1,

=> 2x â€“ 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 âˆˆ 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 âˆˆ 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 âˆˆ R.

Hence proved.

### Question 21. Show that f(x) = x3 â€“ 6x2 + 12x â€“ 18 is increasing on R.

Solution:

We have,

f(x) = x3 â€“ 6x2 + 12x â€“ 18

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x2 â€“ 12x + 12 â€“  0

f'(x) = 3x2 â€“ 12x + 12

f'(x) = 3 (x2 â€“ 4x + 4)

f'(x) = 3 (x â€“ 2)2

Now for x âˆˆ R, we get,

=> (x â€“ 2)2 > 0

=> 3 (x â€“ 2)2 > 0

=> f'(x) > 0

Hence, the function is increasing on the interval x âˆˆ 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 â€“ 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 â€“ 6x + 3

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 2x â€“ 6 + 0

f'(x) = 2x â€“ 6

f'(x) = 2(x â€“ 3)

Now for x âˆˆ [4, 6], we get,

=> 4 â‰¤ x â‰¤ 6

=> 1 â‰¤ (x â€“ 3) â‰¤ 3

=> x â€“ 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 â€“ cos x is an increasing function on (â€“Ï€/4, Ï€/4).

Solution:

We have,

f(x) = sin x â€“ 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 âˆˆ (â€“Ï€/4, Ï€/4)

=> â€“Ï€/4 < x < Ï€/4

=> 0 < (x + Ï€/4) < Ï€/2

=> sin 0 < sin (x + Ï€/4) < sin Ï€/2

=> 0 < sin (x + Ï€/4) < 1

=> sin (x + Ï€/4) > 0

=> âˆš2 sin (x + Ï€/4) > 0

=> f'(x) > 0

Thus, the function is increasing on the interval (â€“Ï€/4, Ï€/4).

Hence proved.

### Question 24. Show that f(x) = tanâ€“1 x â€“ x is a decreasing function on R.

Solution:

We have,

f(x) = tanâ€“1 x â€“ x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

f'(x) =

f'(x) =

Now for x âˆˆ 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 âˆˆ R.

Hence proved.

### Question 25. Determine whether f(x) = â€“x/2 + sin x is increasing or decreasing function on (â€“Ï€/3, Ï€/3).

Solution:

We have,

f(x) = â€“x/2 + sin x

On differentiating both sides with respect to x, we get

f'(x) =

f'(x) =

Now, we have

=> x âˆˆ (â€“Ï€/3, Ï€/3)

=> â€“Ï€/3 < x < Ï€/3

=> cos (â€“Ï€/3) < cos x < cos (Ï€/3)

=> 1/2 < cos x < 1/2

=>  > 0

=> f'(x) > 0

Thus, f(x) is a increasing function on the interval x âˆˆ (â€“Ï€/3, Ï€/3).

Hence proved.

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