If x = a(> 0) divides the area bounded by X–axis, part of the curve y=1+8x2 and the ordinates x = 2, x = 4 into equal parts then a =

A1=A2⇒∫2a 1+8x2dx=∫a4 1+8x2dx ⇒x- 8x2a=x- 8xa4 ⇒a-8a-2-82=4-84-a-8a ⇒2a=16a⇒a=22

If x = a(> 0) divides the area bounded by X–axis, part of the curve y=1+8x2 and the ordinates x = 2, x = 4 into equal parts then a =

A1=A2⇒∫2a 1+8x2dx=∫a4 1+8x2dx ⇒x- 8x2a=x- 8xa4 ⇒a-8a-2-82=4-84-a-8a ⇒2a=16a⇒a=22

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If x = a(> 0) divides the area bounded by X–axis, part of the curve $y = 1 + \frac{8}{x^{2}}$ and the ordinates x = 2, x = 4 into equal parts then a =

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$\frac{1}{\sqrt{2}}$

$2 \sqrt{2}$

$\sqrt{2}$

answer is D.

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Detailed Solution

$A_{1} = A_{2} \Rightarrow \int_{2}^{a} (1 + \frac{8}{x^{2}}) dx = \int_{a}^{4} (1 + \frac{8}{x^{2}}) dx \Rightarrow {(x - \frac{8}{x})}_{2}^{a} = {(x - \frac{8}{x})}_{a}^{4} \Rightarrow (a - \frac{8}{a}) - (2 - \frac{8}{2}) = (4 - \frac{8}{4}) - (a - \frac{8}{a}) \Rightarrow 2 a = \frac{16}{a} \Rightarrow a = 2 \sqrt{2}$