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 $a>0\Rightarrow a=2\sqrt{2}$.