1.

if `int_0^k (dx)/(2+8x^2)=pi/16` then find the value of `k`A. 1B. `(1)/(2)`C. `(1)/(4)`D. None of these

Answer» Correct Answer - B
` int _(0)^(k) 1/(2+8x^(2)) dx = 1/2 int_(0)^(k) (dx)/(1+(2x)^(2))`
`1/4 int _(0)^(2k) (dt)/(1+t^(2)) = 1/4 [ tan^(-1) t ] _(0)^(2k) = 1/4 tan^(-1) 2k `
Comparing it with the given value , we get
` tan ^(-1) 2k = pi/4 rArr 2k = 1 rArr k = 1/2 `


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