UPSC NDA I 2024 Mathematics Paper

Concept:The product of the absolute value function and the greatest integer function is integrated by splitting the interval at the point where the greatest integer function changes value.Explanation:We have $f(x) = |x-1|$ and $g(x) = [x]$, where $[x]$ denotes the greatest integer $\le x$.The function $h(x) = f(x) \cdot g(x)$.Consider the interval $0 < x < 2$.For $0 < x < 1$, we have $[x] = 0$. Therefore $h(x) = |x-1| \cdot 0 = 0$.For $1 < x < 2$, we have $[x] = 1$ and $|x-1| = x-1$ (since $x > 1$). Hence $h(x) = (x-1) \cdot 1 = x-1$.Thus the integral from $0$ to $2$ is the sum of the integrals over the two subintervals:$$\int_0^2 h(x)\,dx = \int_0^1 0\,dx + \int_1^2 (x-1)\,dx.$$Evaluating the second integral:$$\int_1^2 (x-1)\,dx = \left[ \frac{x^2}{2} - x \right]_1^2 = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) = (2-2) - \left( \frac{1}{2} - 1 \right) = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2}.$$Therefore the required integral equals $\frac{1}{2}$.Answer:Option D: $\frac{1}{2}$.