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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 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. unitsf(x)=sinxf(x)+f(π−x)=2f(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/20sinxsindx+∫ππ/2(2−sinx)dx+∫3π/2π(2+sinx)dx+∫2π3π/2(−sinx)dx=1+2×π2−1+2.π2−1=2π sq. units.

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