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

$=\underset{2}{\overset{4}{\int }}ydx$
$=\underset{2}{\overset{4}{\int }}\left(1+\frac{8}{x^2}\right)dx$
$={\left[x-\frac{8}{x}\right]}_2^4=4$
Since, $x=a$ divides this area into two equal parts,
$\therefore$ Required area $=2\underset{2}{\overset{a}{\int }}ydx$
$\therefore 4=2\underset{2}{\overset{a}{\int }}\left(1+\frac{8}{x^2}\right)dx$
$\Rightarrow 2={\left[x-\frac{8}{x}\right]}_2^a=\left(a-\frac{8}{a}\right)-(2-4)$
$\Rightarrow a^2=8$
$\therefore a=2\sqrt{2}$