If the line $x = a$ bisects the area under the curve $y=\frac{1}{x^2}, 1≤x≤9$, then a is equal to
Answer & explanation
Correct answer: option 2
It is given that
$2\int\limits_1^a\frac{1}{x^2}dx=\int\limits_1^9\frac{1}{x^2}dx$
$⇒2\left[-\frac{1}{x}\right]_1^a=\left[-\frac{1}{x}\right]_1^9$
$⇒2\left(-\frac{1}{a}+1\right)=\left(-\frac{1}{9}+1\right)$
$⇒-\frac{2}{a}+2=-\frac{1}{9}+1⇒-\frac{2}{a}=-\frac{10}{9}⇒a=\frac{9}{5}$