Class 12-commerce RD SHARMA Solutions Maths Chapter 17: Increasing and Decreasing Functions
Increasing and Decreasing Functions Exercise MCQ
Solution 31
Given: 
Differentiating w.r.t x, we get
Take f'(x) = 0
For x < -2,
x + 2 < 0 and x + 1 < 0
Therefore, f'(x) > 0
For -2 ≤ x ≤ -1,
x + 2 > 0 and x + 1 < 0
Therefore, f'(x) < 0
Hence, f(x) is decreasing in the interval [-2, -1].
Solution 32
Given: 
Differentiating w.r.t x, we get
Take f'(x) = 0
For x < 1,
x - 3 < 0 and x - 1 < 0
Therefore, f'(x) > 0
For 1 < x < 3,
x - 3 < 0 and x - 1 > 0
Therefore, f'(x) < 0
For x > 3,
x - 3 > 0 and x - 1 > 0
So, f'(x) > 0
Hence, f(x) is decreasing for 1 < x < 3.
Solution 33
Given: 
Differentiating w.r.t x, we get
Now, 
So, f'(x) > 0 when cos x > 0
i.e. 
Therefore, f(x) is increasing when
Also, f'(x) < 0 when cos x < 0
i.e. 
Therefore, f(x) is decreasing when 
Hence, f(x) is decreasing in the interval 
Solution 34
a. Let f(x) = sin 2x
Differentiating w.r.t x, we get
f'(x) = 2cos 2x
When 
Now, 
b. Let g(x) = tan x
Differentiating w.r.t x, we get
f'(x) = sec2x
Now, sec2x > 0 for 
Therefore, f(x) is increasing in the interval 
c. Let h(x) = cos x
Differentiating w.r.t x, we get
f'(x) = -sin x
Now, sin x > 0 for
So, -sin x < 0 for
Therefore, f(x) is decreasing in the interval
d. Let h(x) = cos 3x
Differentiating w.r.t x, we get
f'(x) = -3 sin 3x
When 
Now, sin 3x > 0 for 
Also, sin 3x < 0 for 
Thus, cos x is decreasing in the interval
Solution 35
Given: f(x) = tan x - x
Differentiating w.r.t x, we get
f'(x) = sec2x - 1 = tan2x
Now, tan2x ≥ 0
Hence, f(x) always increases.
Solution 1
Correct option: (b)
Solution 2
Correct option: (c)
Solution 3
Correct option: (c)
Solution 4
Correct option:(b)
Solution 5
Correct option: (a)
Solution 6
Correct option: (c)
Solution 7
Correct option: (b)
Solution 8
Correct option: (c)
Solution 9
Correct option: (a)
Solution 10
Correct option: (c)
Solution 11
Correct option: (b)
Solution 12
Correct option: (a)
Solution 13
Correct option: (b)
Solution 14
Correct option: (d)
Solution 15
Correct option: (d)
Solution 16
Correct option: (c)
Solution 17
Correct option: (c)
Solution 18
Correct option: (d)
Solution 19
Correct option: (a)
Every invertible function is always monotonic function.
Solution 20
Correct option: (b)
Solution 21
Correct option: (c)
Solution 22
Correct option: (a)
Solution 23
Correct option: (d)
Solution 24
Correct option: (d)
Solution 25
Correct option: (b)
Solution 26
Correct option: (b)
Solution 27
Correct option: (d)
Solution 28
Correct option: (a)
Solution 29
Correct option: (a)
NOTE: Option (a) should be -1 < k < 1.
Solution 30
Correct option: (a)
Increasing and Decreasing Functions Exercise Ex. 17.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Increasing and Decreasing Functions Exercise Ex. 17.2
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 1(viii)
Solution 1(ix)
Solution 1(xi)
Solution 1(xii)
Solution 1(xiii)
Solution 1(xiv)
Solution 1(xv)
Solution 1(xvi)
Solution 1(xvii)
Solution 1(xviii)
Solution 1(xix)
Solution 1(xx)
Solution 1(xxi)
Solution 1(xxii)
Solution 1(xxiii)
Solution 1(xxiv)
Solution 1(xxv)
Solution 1(xxvi)
Solution 1(xxvii)
Solution 1(xxviii)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30(i)
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39(i)
Solution 39(ii)
Solution 39(iii)
Solution 1(x)
Given: 
Differentiating w.r.t x, we get
Take f'(x) = 0
Clearly, f'(x) > 0 if x < -2 or x > -1
And, f'(x) < 0 if -2 < x < -1
Thus, f(x)
increases on 
and decreases
on 
Solution 1(xxix)
Given: 
Differentiating w.r.t x, we get
Take f'(x) = 0
The points x = 2,
4 and -3 divide the number line into four disjoint intervals namely 
Consider the
interval 
In this case, x - 2 < 0, x - 4 < 0 and x + 3 < 0
Therefore, f'(x)
< 0 when 
Thus the function
is decreasing in 
Consider the
interval 
In this case, x - 2 < 0, x - 4 < 0 and x + 3 > 0
Therefore, f'(x)
> 0 when 
Thus the function
is increasing in 
Now, consider the
interval 
In this case, x - 2 > 0, x - 4 < 0 and x + 3 > 0
Therefore, f'(x)
< 0 when 
Thus the function
is decreasing in 
And now, consider
the interval 
In this case, x - 2 > 0, x - 4 > 0 and x + 3 > 0
Therefore, f'(x)
< 0 when 
Thus the function
is increasing in 
Solution 30(ii)
Given: 
Differentiating w.r.t x, we get
Now, 
Hence, f(x) is an increasing function for all x.
Increasing and Decreasing Functions Exercise Ex.17VSAQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
