Class 12 RD Sharma Solutions â Chapter 17 Increasing and Decreasing Functions â Exercise 17.2 | Set 1
Question 1. Find the intervals in which the following functions are increasing or decreasing.
(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 < â6/4
=> x < â3/2
=> x â (ââ, â3/2)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> â 6 â 4x < 0
=> â 4x < 6
=> x > â6/4
=> x > â3/2
=> x â ( â3/2, â)
Thus, f(x) is increasing on the interval x â (ââ, â3/2) and decreasing on the interval x â ( â3/2, â).
(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 > â2
=> x > â2/2
=> x > â1
=> x â (â1, â)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> 2x + 2 < 0
=> 2x < â2
=> x < â2/2
=> x < â1
=> x â (ââ, â1)
Thus, f(x) is increasing on the interval x â (â1, â) and decreasing on the interval x â ( ââ, â1).
(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
=> â9 â 2x > 0
=> â2x > 9
=> x > â9/2
=> x â (â9/2, â)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> â9 â 2x < 0
=> â2x < 9
=> x < â9/2
=> x â (ââ, â9/2)
Thus, f(x) is increasing on the interval x â (â9/2, â) and decreasing on the interval x â ( ââ, â9/2).
(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 â (ââ, 1)⪠(3, â) and decreasing on the interval x â (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 (âx2 + 3x â 2x + 6) = 0
=> âx2 + 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 â2 < x < 3.
Also fâ(x) < 0 if x < â2 and x > 3.
Thus, f(x) is increasing on x â (â2, 3) and f(x) is decreasing on interval x â (ââ, â2) ⪠(3, â).
(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 (âx2 + 3x â 2x + 6) = 0
=> âx2 + 3x â 2x + 6 = 0
=> x2 â 3x + 2x â 6 = 0
=> (x â 3) (x + 2) = 0
=> x = 3, â2
Clearly, fâ(x) > 0 if â2 < x < 3.
Also fâ(x) < 0 if x < â2 and x > 3.
Thus, f(x) is increasing on x â (â2, 3) and f(x) is decreasing on interval x â (ââ, â2) ⪠(3, â).
(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, â2
Clearly, fâ(x) > 0 if x < â2 and x > 4.
Also fâ(x) < 0 if â2 < x < 4.
Thus, f(x) is increasing on x â (ââ,â2) ⪠(4, â) and f(x) is decreasing on interval x â (â2, 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, â2
Clearly, fâ(x) > 0 if x < â2 and x > 6.
Also fâ(x) < 0 if â2< x < 6
Thus, f(x) is increasing on x â (ââ,â2) ⪠(6, â) and f(x) is decreasing on interval x â (â2, 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 â (ââ, 2) ⪠(3, â) and f(x) is decreasing on interval x â (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 = â1, â2
Clearly, fâ(x) > 0 if â2 < x < â1.
Also fâ(x) < 0 if x < â1 and x > â2.
Thus, f(x) is increasing on x â (â2,â1) and f(x) is decreasing on interval x â (ââ, â2) ⪠(â2, â).
(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 â (ââ, 1) ⪠(2, â) and f(x) is decreasing on interval x â (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 (âx2 + x + 2) = 0
=> x2 â x â 2 = 0
=> x2 â 2x + x â 2 = 0
=> (x â 2) (x + 1) = 0
=> x = 2, â1
Clearly, fâ(x) > 0 if â1 < x < 2.
Also fâ(x) < 0 if x < â1 and x > 2.
Thus, f(x) is increasing on x â (â1, 2) and f(x) is decreasing on interval x â (ââ, â1) ⪠(2, â).
(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, â2
Clearly, fâ(x) > 0 if x < â2 and x > 2.
Also fâ(x) < 0 if â2 < x < 2.
Thus, f(x) is increasing on x â (ââ, â2) ⪠(2, â), and f(x) is decreasing on interval x â (â2, 2).
(xiv) f(x) = â2x3 â 9x2 â 12x + 1
Solution:
We are given,
f(x) = â2x3 â 9x2 â 12x + 1
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) = â6x2 â 18x â 12 + 0
f'(x) = â6x2 â 18x â 12
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> â6x2 â 18x â 12 = 0
=> 6 (âx2 â 3x â 2) = 0
=> x2 + 3x + 2 = 0
=> x2 + 2x + x + 2 = 0
=> (x + 2) (x + 1) = 0
=> x = â2, â1
Clearly, fâ(x) > 0 if x < â1 and x > â2.
Also, fâ(x) < 0 if â2 < x < â1.
Thus, f(x) is increasing on x â (â2, â1) and f(x) is decreasing on interval x â (ââ, â2) ⪠(â1, â).
(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 â (ââ, 4/3) ⪠(2, â) and f(x) is decreasing on interval x â (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 â (ââ, 2) ⪠(6, â) and f(x) is decreasing on interval x â (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, â2
Clearly, fâ(x) > 0 if x < â2 and x > 2.
Also fâ(x) < 0 if â2 < x < 2.
Thus, f(x) is increasing on x â (ââ, â2) ⪠(2, â), and f(x) is decreasing on interval x â (â2, 2).
(xviii) f(x) = 3x4/10 â 4x3/5 -3x2 + 36x/5 + 11
Solution:
We are given,
f(x) = 3x4/10 â 4x3/5 -3x2 + 36x/5 + 11
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) = 6x3/5 â 12x2/5 -3(2x) + 36/5
f'(x) = 6/5[(x â 1)(x + 2)(x â 3)]
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6/5[(x â 1)(x + 2)(x â 3)] = 0
=> x = 1, â2, 3
Clearly, fâ(x) > 0 if â2 < x < 1 and if x > 3
Also fâ(x) < 0 if 1 < x < 3.
Thus, f(x) is increasing on x â (3, â) and f(x) is decreasing on interval x â (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 â (1, â), and f(x) is decreasing on interval x â (ââ, 1).
(xx) f(x) = x4/4 + 2/3x3 â 5/2x2 â 6x + 7
Solution:
We have,
f(x) = x4/4 + 2/3x3 â 5/2x2 â 6x + 7
On differentiating both sides with respect to x, we get
f'(x) =
f'(x) = 4x3/4 + 6x2/3 â 10x/2 â 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 = â1, â3, 2
Clearly f'(x) > 0 if â3 < x < â1 and x > 2.
Also f'(x) < 0 if x < â3 and â1 < x < 2.
Thus, f(x) is increasing on x â (â3, â1) ⪠(2, â) and f(x) is decreasing on interval x â (ââ, â3) ⪠(â1, 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 â (0, 1) ⪠(2, â) and f(x) is decreasing on interval x â (ââ, 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 â 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 â (0, 1) and f(x) is decreasing on interval x â (1, â).
(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 â (0, â) and f(x) is decreasing on interval x â (ââ, 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 â (ââ, 1) ⪠(3, â) and f(x) is decreasing on interval x â (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 â (0, 1) ⪠(2, â) and f(x) is decreasing on interval x â (ââ, 0) ⪠(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, â1, 2
Clearly f'(x) > 0 if â1 < x < 0 and x > 2.
Also f'(x) < 0 if x < â1 and 0< x < 2.
Thus, f(x) is increasing on x â (â1, 0) ⪠(2, â) and f(x) is decreasing on interval x â (ââ, â1) ⪠(0, 2).
(xxvii) f(x) = 3x4/2 â 4x3 â 45x2 + 51
Solution:
We have,
f(x) = 3x4/2 â 4x3 â 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, â3, 5
Clearly f'(x) > 0 if â3 < x < 0 and x > 5.
Also f'(x) < 0 if x < â3 and 0< x < 5.
Thus, f(x) is increasing on x â (â3, 0) ⪠(5, â) and f(x) is decreasing on interval x â (ââ, â3) ⪠(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 â (2, â) and f(x) is decreasing on interval x â (ââ, 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, â) and f(x) is decreasing on interval x â (ââ, 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 = â1
=> 2x â 6 = â1
=> 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 < 2Ď 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 = â1
=> x = 3Ď/4 , 7Ď/4
Clearly f'(x) > 0 if 0 < x < 3Ď/4 and 7Ď/4 < x < 2Ď.
Also f'(x) < 0 if 3Ď/4 < x < 7Ď/4.
Thus, f(x) is increasing on x â (0, 3Ď/4) ⪠(7Ď/4, 2Ď) and f(x) is decreasing on interval x â (3Ď/4, 7Ď/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 â 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, Ď/2) and decreasing on (Ď/2, Ď) and neither increasing nor decreasing in (0, Ď).
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 < Ď/2,
=> cos x > 0
=> f'(x) > 0
And for Ď/2 < x < Ď,
=> cos x < 0
=> f'(x) < 0
Thus, f(x) is increasing on x â (0, Ď/2) and f(x) is decreasing on interval x â (Ď/2, Ď).
Hence f(x) is neither increasing nor decreasing in (0, Ď).
Hence proved.
Question 8. Show that f(x) = log sin x is increasing on (0, Ď/2) and decreasing on (Ď/2, Ď).
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 < Ď/2,
=> cot x > 0
=> f'(x) > 0
And for Ď/2 < x < Ď,
=> cos x < 0
=> f'(x) < 0
Thus, f(x) is increasing on x â (0, Ď/2) and f(x) is decreasing on interval x â (Ď/2, Ď).
Hence proved.
Question 9. Show that f(x) = x â sin x is increasing for all x â 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 â R, we get
=> â1 < cos x < 1
=> â1 > cos x > 0
=> fâ(x) > 0
Thus, f(x) is increasing on interval x â R.
Hence proved.
Question 10. Show that f(x) = x3 â 15x2 + 75x â 50 is an increasing function for all x â 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 Ďľ 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 â R.
Hence proved.