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Question

# The values of ′a′ and ′b′ so that the functionf(x)= ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x+a√2sinx,0≤x<π42xcotx+b,π4≤x≤π2acos2x−bsinx,π2<x≤π is continuous at x=π4,π2 are

A
a=π6
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B
a=π6
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C
b=π12
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D
b=π
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Solution

## The correct option is C b=−π12f(x) is continuous in the interval 0≤x<π4,π4<x<π2,π2<x≤π We need to make the function continuous at x=π4,π2 For continuity at x=π4,limx→π4−f(x)=limx→π4+f(x)=f(π4) limx→(π4)−(x+a√2sinx)=limx→(π4)+(2xcotx+b)=f(π4) ⇒π4+a√2sin(π4)=2.π4.cot(π4)+b=2π4.cot(π4)+b ⇒π4+a=π2+b⇒a−b=π4 ...(1) For continuity at x=π2,limx→π2−f(x)=limx→π2+f(x)=f(π2) ⇒limx→π2−(2xcotx+b)=limx→(π2)+(acos2x−bsinx)=acosπ−bsinπ2 ⇒0+b=−a−b⇒a+2b=0 ...(2) From equation (1) and (2) a=π6,b=−π12

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