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Question

If f(x)=sin1(cosx)cos1(sinx) x[0,2π], then which of the following statement(s) is/are correct?

A
f(x) is differentiable in x[0,2π]
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B
Range of f(x) is [π24,π24]
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C
x=π is a point of global minima as well as local minima.
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D
π0f(x) dx=0
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Solution

The correct options are
B Range of f(x) is [π24,π24]
C x=π is a point of global minima as well as local minima.
D π0f(x) dx=0
Given : f(x)=sin1(cosx)cos1(sinx)
f(x)=(π2cos1(cosx))(π2sin1(sinx))f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪(π2x)(π2x) ; 0xπ2(π2x)(π2(πx)) ; π2xπ(π2(2πx))(π2(πx)) ; πx<3π2(π2(2πx))(π2(x2π)) ; 3π2x2π

f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪(π2x)2 ; 0xπ2(π2x)2 ; π2xπ(πx)2π24 ; πx<3π2π24(2πx)2 ; 3π2x2π

f(x) is non differentiable at x=π
Range of f(x) is [π24,π24]
x=π is a point of global minima as well as local minima.

Now,
π0f(x) dx=π/20(π2x)2 dxππ/2(π2x)2 dx=0

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