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Class 12-science 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

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