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Question

If f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪1sinx(π2x)2logsinxlog(1+π24πx+x2),xπ2k,x=π2 is continuous at x=π2, then k is equal to.

A
116
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B
132
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C
164
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D
128
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Solution

The correct option is D 164
For f(x) to be continuous at x=π2,
We must have
limxπ/2f(x)=f(π/2)
limxπ/21sinx(π2x)2log(sinx)log(1+π24πx+4x2)=k

limh01cosh4h2×logcoshlog(1+4h2)=k .... [x=π2hh0 as xπ2]

limh01cosh4h2×log{1+cosh1}cosh1×4h2log(1+4h2)×cosh14h2=k

limh0(1cosh4h2)2log(1+(cosh1))cosh1×4h2log(1+4h2)=k
limh0(sin2h/22h2)2log(1+(cosh1))cosh1×4h2log(1+4h2)=k
164limh0(sinh/2h/2)4log(1+(cosh1))cosh1×4h2log(1+4h2)=k
164=k.

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