UPSC NDA I 2016 Mathematics Solved Paper

Concept:The integral of a product involving the greatest integer function requires splitting the interval where the integer part is constant.Explanation:We are given $f(x) = x \, g(x)$ with $g(x) = \left[ \frac{1}{x} \right]$, the greatest integer function.Thus $f(x) = x \left[ \frac{1}{x} \right]$.The integral is $I = \int_{1/3}^{1} x \left[ \frac{1}{x} \right] dx$.For $x$ in $(1/3, 1/2)$, we have $\frac{1}{x}$ between $2$ and $3$, so $\left[ \frac{1}{x} \right] = 2$.For $x$ in $(1/2, 1)$, we have $\frac{1}{x}$ between $1$ and $2$, so $\left[ \frac{1}{x} \right] = 1$.Hence $I = \int_{1/3}^{1/2} 2x \, dx + \int_{1/2}^{1} x \, dx$.Compute: $\int 2x\,dx = x^2$, so first part $= \left[ x^2 \right]_{1/3}^{1/2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.Second part: $\int x\,dx = \frac{x^2}{2}$, so $= \frac{1}{2} \left[ x^2 \right]_{1/2}^{1} = \frac{1}{2} \left(1 - \frac{1}{4}\right) = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$.Total $I = \frac{5}{36} + \frac{3}{8} = \frac{10}{72} + \frac{27}{72} = \frac{37}{72}$.Answer:$\frac{37}{72}$ (Option A)