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Q.

 Let f(x)=-2sinx  -πx-π2asinx+b -π2<x<π2cosx π2xπ If f is continuous on [-π,π), then 

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a

b=1, a=1

b

b=1, a=-1

c

None

d

b=-1, a=-1

answer is A.

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Detailed Solution

The given function is f(x)=-2sinx  -πx-π2asinx+b -π2<x<π2cosx π2xπ

Since f(x) s continuous on (-π,π)

f is continuous at x=-π2 and at x=π2

limx-π2-f(x)=limx-π2+fx

-2sin-π2=asin-π2+b

2=b-a...1

And limxπ2-f(x)=limxπ2+fx

asinπ2+b=cosπ2 a+b=0. a=-b...2

From equations 1, 2 we get, b=1, a=-1

Hence option-1 is the correct answer.

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 Let f(x)=-2sinx  -π≤x≤-π2asinx+b -π2<x<π2cosx π2≤x≤π If f is continuous on [-π,π), then