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 twice equal parts, then $a=$

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

Answer 1

The correct option is B $2\sqrt{2}$

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

${\int }_4^2{(1+\frac{8}{x^2})}^{dx}=2{\int }_4^2(1+\frac{8}{x^2})dx$

${[x-\frac{8}{x}]}_2^4=2\times {[x-\frac{8}{x}]}_2^4$

$\left[4\frac{8}{4}\right]-[2-\frac{8}{2}]=2[4-\frac{8}{4}]-2[a-\frac{8}{a}]$

$2-(-2)=4-2[a-\frac{8}{a}]$

$[a-\frac{8}{a}]=0\Rightarrow a=\frac{8}{a}\to a^2=8\Rightarrow a=2\sqrt{2}$

Ignoring $-2\sqrt{2}$ as $2\sqrt{2}$ comes between $2$