If -0k dx-2-8x2-16-then k-

∫_0^k 1/2+8x^2dx=1/2∫_0^k dx/1+(2x)^2=1/4∫_0^2kdt/1+t^2 =1/4|tan^ 1t|_0^2k=1/4tan^ 12k. Comparing it with the given value, we get tan^ 12k=π/4⇒ 2k=1⇒ k=1/2 ...

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Standard XIII

Mathematics

Property 1

If ∫0k dx/2+8...

Question

$I f \int_{0}^{k} \frac{d x}{2 + 8 x^{2}} = \frac{π}{16}, t h e n k =$

1

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\frac{1}{2}

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\frac{1}{4}

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N o n e o f t h e s e

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Solution

The correct option is B $\frac{1}{2}$
$\int_{0}^{k} \frac{1}{2 + 8 x^{2}} d x = \frac{1}{2} \int_{0}^{k} \frac{d x}{1 + (2 x)^{2}} = \frac{1}{4} \int_{0}^{2 k} \frac{d t}{1 + t^{2}}$
$= \frac{1}{4} | t a n^{- 1} t |_{0}^{2 k} = \frac{1}{4} t a n^{- 1} 2 k .$
Comparing it with the given value, we get $t a n^{- 1} 2 k = \frac{π}{4} \Rightarrow 2 k = 1 \Rightarrow k = \frac{1}{2}$

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If ∫k0dx2+8x2=π16,then k=

The correct option is B 12∫k012+8x2dx=12∫k0dx1+(2x)2=14∫2k0dt1+t2 =14|tan−1t|2k0=14tan−12k. Comparing it with the given value, we get tan−12k=π4⇒2k=1⇒k=12

$I f \int_{0}^{k} \frac{d x}{2 + 8 x^{2}} = \frac{π}{16}, t h e n k =$