Question

Let the area bounded by the $x$ -axis, curve $y=\left(1+\frac{8}{x^2}\right)$ and the ordinates $x=2$ and $x=4$ is ''$A$'' sq. unit and if the ordinate $x=a$ divides the area into two equal parts, then the correct statement among the following is

• A. $A=3$ and $a=2$
• B. $A=2$ and $a=1$
• C. $A=4\sqrt{2}$ and $a=\sqrt{2}$
• D. $A=4$ and $a=2\sqrt{2}$

Answer 1

Answer: (D)

$y=1+\frac{8}{x^2}$


Required area $A=\underset{2}{\overset{4}{\int }}ydx=\underset{2}{\overset{4}{\int }}\left(1+\frac{8}{x^2}\right)dx$
$={\left[x-\frac{8}{x}\right]}_2^4=4$
If $x=a$ bisects the area then we have $\underset{2}{\overset{a}{\int }}\left(1+\frac{8}{x^2}\right)=\frac{4}{2}\dots \left(1\right)$
Now $\underset{2}{\overset{a}{\int }}\left(1+\frac{8}{x^2}\right)dx={\left[x-\frac{8}{x}\right]}_2^a=\left[a-\frac{8}{a}-2+4\right]$
Now $a-\frac{8}{a}+2=2$ (by equation (1))
$\Rightarrow a-\frac{8}{a}=0\Rightarrow a^2=8$
$\Rightarrow a=\pm 2\sqrt{2}$
Since $2<a<4$
$\therefore a=2\sqrt{2}$