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Question

# If f(x)=sinx,∀ x ∈[0,π2], f(x)+f(π−x)=2, ∀ x∈(π2,π] and f(x)=f(2π−x), ∀ x∈(π,2π], then the area enclosed by y=f(x) and the x-axis is

A
π sq. units
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B
2π sq. units
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C
2 sq. units
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D
4 sq. units
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Solution

## The correct option is B 2π sq. units f(x)=sinx f(x)+f(π−x)=2 f(x)=2−f(π−x)=2−sin(π−x)=2−sinx, where x∈(π2,π] f(x)=f(2π−x)=2−sin(2π−x), where x∈(π,3π2] f(x)=f(2π−x)=−sinx, where x∈(3π2,2π] f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩sinx,x∈[0,π2]2−sinx,x∈(π2,π]2+sinx,x∈(π,3π2]−sinx,x∈(3π2,2π] Area =π/2∫0sinx dx+π∫π/2(2−sinx)dx+ 3π/2∫π(2+sinx)dx+2π∫3π/2(−sinx)dx =1+2×π2−1+2×π2−1+1 =2π sq. units

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