Class 12 RD Sharma Solutions â Chapter 17 Increasing and Decreasing Functions â Exercise 17.2 | Set 2
Question 11. Show that f(x) = cos2 x is a decreasing function on (0, Ď/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 < Ď/2,
=> sin 2x > 0
=> â sin 2x < 0
=> f'(x) < 0
Thus, f(x) is decreasing on x â (0, Ď/2).
Hence proved.
Question 12. Show that f(x) = sin x is an increasing function on (âĎ/2, Ď/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 âĎ/2 < x < Ď/2,
=> cos x > 0
=> f'(x) > 0
Thus, f(x) is increasing on x â (âĎ/2, Ď/2).
Hence proved.
Question 13. Show that f(x) = cos x is a decreasing function on (0, Ď), increasing in (âĎ, 0) and neither increasing nor decreasing in (âĎ, Ď).
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 < Ď,
=> sin x > 0
=> â sin x < 0
=> fâ(x) < 0
And for âĎ < x < 0,
=> sin x < 0
=> â sin x > 0
=> fâ(x) > 0
Therefore, f(x) is decreasing in (0, Ď) and increasing in (âĎ, 0).
Hence f(x) is neither increasing nor decreasing in (âĎ, Ď).
Hence proved.
Question 14. Show that f(x) = tan x is an increasing function on (âĎ/2, Ď/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 âĎ/2 < x < Ď/2,
=> sec2 x > 0
=> fâ(x) > 0
Thus, f(x) is increasing on interval (âĎ/2, Ď/2).
Hence proved.
Question 15. Show that f(x) = tanâ1 (sin x + cos x) is a decreasing function on the interval (Ď/4, Ď /2).
Solution:
We have,
f(x) = tanâ1 (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 Ď/4 < x < Ď/2,
=> < 0
=> fâ(x) < 0
Thus, f(x) is decreasing on interval (Ď/4, Ď/2).
Hence proved.
Question 16. Show that the function f(x) = sin (2x + Ď/4) is decreasing on (3Ď/8, 5Ď/8).
Solution:
We have,
f(x) = sin (2x + Ď/4)
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) = 2 cos (2x + Ď/4)
Now we have, 3Ď/8 < x < 5Ď/8
=> 3Ď/4 < 2x < 5Ď/4
=> 3Ď/4 + Ď/4 < 2x + Ď/4 < 5Ď/4 + Ď/4
=> Ď < 2x + Ď/4 + 3Ď/2
As, 2x + Ď/4 lies in 3rd quadrant, we get,
=> cos (2x + Ď/4) < 0
=> 2 cos (2x + Ď/4) < 0
=> f'(x) < 0
Thus, f(x) is decreasing on interval (3Ď/8, 5Ď/8).
Hence proved.
Question 17. Show that the function f(x) = cotâ1 (sin x + cos x) is increasing on (0, Ď/4) and decreasing on (Ď/4, Ď/2).
Solution:
We have,
f(x) = cotâ1 (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 Ď/4 < x < Ď/2,
=> < 0
=> cos x â sin x < 0
=> fâ(x) < 0
Also for 0 < x < Ď/4,
=> > 0
=> cos x â sin x > 0
=> f'(x) > 0
Thus, f(x) is increasing on interval (0, Ď/4) and decreasing on intervals (Ď/4, Ď/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 â 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 â 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 â 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 â 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 â 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 â [4, 6], we get,
=> 4 ⤠x ⤠6
=> 1 ⤠(x â 3) ⤠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 (âĎ/4, Ď/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 â (âĎ/4, Ď/4)
=> âĎ/4 < x < Ď/4
=> 0 < (x + Ď/4) < Ď/2
=> sin 0 < sin (x + Ď/4) < sin Ď/2
=> 0 < sin (x + Ď/4) < 1
=> sin (x + Ď/4) > 0
=> â2 sin (x + Ď/4) > 0
=> f'(x) > 0
Thus, the function is increasing on the interval (âĎ/4, Ď/4).
Hence proved.
Question 24. Show that f(x) = tanâ1 x â x is a decreasing function on R.
Solution:
We have,
f(x) = tanâ1 x â x
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) =
f'(x) =
f'(x) =
Now for x â 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 â R.
Hence proved.
Question 25. Determine whether f(x) = âx/2 + sin x is increasing or decreasing function on (âĎ/3, Ď/3).
Solution:
We have,
f(x) = âx/2 + sin x
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) =
Now, we have
=> x â (âĎ/3, Ď/3)
=> âĎ/3 < x < Ď/3
=> cos (âĎ/3) < cos x < cos (Ď/3)
=> 1/2 < cos x < 1/2
=> > 0
=> f'(x) > 0
Thus, f(x) is a increasing function on the interval x â (âĎ/3, Ď/3).
Hence proved.