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Q.

∫0k dx2+8x2=π16 then k=

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a

0

b

12

c

π2

d

2

answer is B.

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Detailed Solution

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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∫0k dx2+8x2=π16 then k=