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

  • Last Updated : 02 Jun, 2021

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

Solution:

We have,

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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,

=> sec2 x > 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

Noe 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.




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