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A function f is defined by f(x)=int0^pi ...

A function `f` is defined by `f(x)=int_0^pi costcos(x-t)dt, 0lexle2pi`. Which of the following hold(s) good? (A) `f(x)` is continuous but not differentiable in `(0,2pi)` (B) There exists at least one `c in (0,2pi)` such that `f\'(c)=0` (C) Maximum value of `f` is `pi/2` (D) Minimum value of `f` is `-pi/2`

A

f(x) is continuous but not differentiable in `(0, 2pi)`.

B

Maximum value of f is `pi//2`

C

There exists atleast one `c in (0, 2pi)` such that `f'(c)=0`

D

Minimum value of f is `-(pi)/(2)`.

Text Solution

Verified by Experts

The correct Answer is:
B, C, D
`f(x)=int_(0)^(pi)costcos(x-t)dt" (i)"`
`=int_(0)^(pi)-cost cos(x-pi_t)dt`
`therefore" "f(x)=int_(0)^(pi)cost.cos(x+t)dt`
Adding (i) and (ii), we get
`2f(x)=int_(0)^(pi)cost(2cosx. cost)dt`
`therefore" "f(x)=cosx int_(0)^(pi)cos^(2)tdt=2cosx int_(0)^(pi//2)cos^(2)tdt`
`therefore" "f(x)=(picosx)/(2)`, which is continuous and differentiable,
having maximum value `pi//2` and minimum value `-pi//2`.
Also f(x) satisfies all the conditions of Rolle's theorem in `[0,2pi]`.
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