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Question

Let f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪2sinx,πxπ2asinx+b,π2<x<π2cosx,π2xπ
If f(x) is continuous on [π,π], then

A
a=1,b=1
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B
a=1,b=1
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C
a=1b=1
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D
a=1,b=1
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Solution

The correct option is C a=1b=1
Given f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪2sinx,πxπ2asinx+b,π2<x<π2cosx,π2xπ

Since f(x) is continuous on [π,π], then

limxπ2f(x)=limxπ+2f(x)

limxπ22sinx=limxπ2asinx+b

2sin(π2)=asin(π2)+b

2=a+b...(i)

and limxπ2f(x)=limxπ+2f(x)

limxπ2asinx+b=limxπ2cosx

asinπ2+b=cosπ2

a+b=0...(ii)

On solving equations (i) and (ii), we get

a=1 and b=1

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