Q. 123.9( 38 Votes )

# Which of the foll

Answer :

(A) Let f_{1}(x) = cosx

In interval,

Therefore, f_{1}(x) = cosx is strictly decreasing in interval.

(B) Let f_{2}(x) = cos2x

Now, 0 < x <

⇒ 0 < 2x < π

⇒ sin2x > 0

⇒ -2sin2x < 0

Therefore, f_{2}(x) = cos2x is strictly decreasing in interval.

(C) Let f_{3}(x) = cos3x

Now,

⇒ sin3x = 0

⇒ 3x = π, as xϵ

⇒ x =

The point x = divides the interval into two distinct intervals.

i.e. and

Now, in interval, ,

f_{3}'(x) = -3sin3x < 0 as (0 < x < => 0 < 3x < π)

Therefore, f_{3} is strictly decreasing in interval.

Now, in interval

f_{3}'(x)=-3sin3x > 0 as

Therefore, f_{3} is strictly increasing in interval.

(D) Let f_{4} = tanx

In interval,

Therefore, f4 is strictly increasing in interval .

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Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

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Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1

Find the intervalRD Sharma - Volume 1