The value of $$a\left( {a > 0} \right)$$ 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 :
$$1$$
Solution :
$$\eqalign{
& f\left( a \right) = \int\limits_a^{2a} {\left( {\frac{x}{6} + \frac{1}{{{x^2}}}} \right)dx} \cr
& = \left( {\frac{{{x^2}}}{{12}} - \frac{1}{x}} \right)_a^{2a} \cr
& = \left( {\frac{{4{a^2}}}{{12}} - \frac{1}{{2a}} - \frac{{{a^2}}}{{12}} + \frac{1}{a}} \right) \cr
& = \frac{{{a^2}}}{4} + \frac{1}{{2a}} \cr
& {\text{Let }}f'\left( a \right) = \frac{{2a}}{4} - \frac{1}{{2{a^2}}} = 0 \cr
& \Rightarrow a = 1{\text{ which is a point of minima}}{\text{.}} \cr} $$
Releted MCQ Question on Calculus >> Application of Integration
Releted Question 1
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-