Question

The value of $a,(a>0)$ for which the area bounded by the curves $y=\frac{x}{6}+\frac{1}{x^2},y=0,x=a$ and $x=2a$ has the least value is

• A. $2$
• B. $\sqrt{2}$
• C. $2^{\frac{1}{3}}$
• D. $1$

Answer 1

The correct option is D $1$

$A={\int }_a^{2a}(\frac{x}{6}+\frac{1}{x^2})dx={[\frac{x^2}{12}-\frac{1}{x}]}_a^{2a}=\frac{a^2}{4}+\frac{1}{2a}$
Now,

$f\left(a\right)=\frac{a^2}{4}+\frac{1}{2a}$

Differentiating both sides, we get

$\Rightarrow f^{\prime }\left(a\right)=\frac{a}{2}-\frac{1}{2a^2}$

For minimum value, $f^{\prime }\left(a\right)=0$
$\Rightarrow a=1$
$\because f"\left(a\right)>0$; so, at $a=1,f\left(a\right)$ is minimum.