Question

If the ordinate $x=a$ divides the area bounded by the curve $y=(1+\frac{8}{x^2}),$ x - axis and the ordinates $x=2,x=4$ into two equal parts, then a =

• A. 8
• B. $2\sqrt{2}$
• C. 2
• D. $\sqrt{2}$

Answer 1

The correct option is B

$2\sqrt{2}$

Let the ordinate at $x=a$ divide the area into two equal parts

Area of $AMNB={\int }_2^4(1+\frac{8}{x^2})dx={[x-\frac{8}{x}]}_2^4=4$
Area of $ACDM={\int }_2^a(1+\frac{8}{x^2})dx=2$
On solving, we get $a=\pm 2\sqrt{2}$; Since a > 0 $\Rightarrow$ a = 2$\sqrt{2}$