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0kdx2+8x2=π16 then k=

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a
0
b
12
c
π2
d
2

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detailed solution

Correct option is B

G.I =12∫0K dx1+4x2=12∫0k dx1+(2x)2=12⋅12⋅tan−1⁡(2x]0k⇒14tan−1⁡(2k)=π16tan−1⁡(2k)=π42k=1k=12


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