The value of a (a 0) for which the area bou | vedclass

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(D) The area $A$ is given by the integral: $A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx$ $= \left[ \frac{x^2}{12} - \frac{1}{x} \r

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$1$

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(D) The area $A$ is given by the integral: $A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx$ $= \left[ \frac{x^2}{12} - \frac{1}{x} \right]_{a}^{2a}$ $= \left( \frac{(2a)^2}{12} - \frac{1}{2a} \right) - \left( \frac{a^2}{12} - \frac{1}{a} \right)$ $= \left( \frac{4a^2}{12} - \frac{1}{2a} \right) - \left( \frac{a^2}{12} - \frac{1}{a} \right)$ $= \frac{3a^2}{12} + \frac{1}{a} - \frac{1}{2a} = \frac{a^2}{4} + \frac{1}{2a}$ Let $f(a) = \frac{a^2}{4} + \frac{1}{2a}$. To find the minimum,we differentiate with respect to $a$: $f'(a) = \frac{2a}{4} - \frac{1}{2a^2} = \frac{a}{2} - \frac{1}{2a^2}$ Setting $f'(a) = 0$: $\frac{a}{2} = \frac{1}{2a^2} \Rightarrow a^3 = 1 \Rightarrow a = 1$ Thus,the value of $a$ for which the area is minimum is $1$.

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

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