Question

If the ordinate $x=a$ 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 two equal parts, then $a$ is equal to

• A. $\sqrt{2}$
• B. $2 \sqrt{2}$
• C. $3 \sqrt{2}$
• D. None of these

Answer 1

Answer: (B)

The area bounded by the curve $y=1+\frac{8}{x^{2}}, x$ - axis and the ordinates $x=2, x=4$ is

$=\int\limits_{2}^{4} y d x$
$=\int\limits_{2}^{4}\left(1+\frac{8}{x^{2}}\right) d x$
$=\left[x-\frac{8}{x}\right]_{2}^{4}=4$
Since, $x=a$ divides this area into two equal parts,
$\therefore $ Required area $=2 \int\limits_{2}^{a} y d x $
$\therefore 4=2 \int\limits_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x$
$\Rightarrow 2=\left[x-\frac{8}{x}\right]_{2}^{a}=\left(a-\frac{8}{a}\right)-(2-4) $
$\Rightarrow a^{2}=8$
$ \therefore a=2 \sqrt{2}$