    Question

# If f(x)=sin−1(cosx)cos−1(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 π∫0f(x) dx=0Given : f(x)=sin−1(cosx)cos−1(sinx) ⇒f(x)=(π2−cos−1(cosx))(π2−sin−1(sinx))⇒f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(π2−x)(π2−x) ; 0≤x≤π2(π2−x)(π2−(π−x)) ; π2≤x≤π(π2−(2π−x))(π2−(π−x)) ; π≤x<3π2(π2−(2π−x))(π2−(x−2π)) ; 3π2≤x≤2π ⇒f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(π2−x)2 ; 0≤x≤π2−(π2−x)2 ; π2≤x≤π(π−x)2−π24 ; π≤x<3π2π24−(2π−x)2 ; 3π2≤x≤2π 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, π∫0f(x) dx=π/2∫0(π2−x)2 dx−π∫π/2(π2−x)2 dx=0  Suggest Corrections  0      Similar questions  Explore more