UPSC NDA II 2018 Mathematics Solved Paper

Concept:The absolute value function splits at the point where the expression inside becomes zero.For $|x - 5|$, the split point is $x = 5$.The integral over $[2, 8]$ is broken into two parts: $[2, 5]$ and $[5, 8]$.Explanation:First, rewrite $|x - 5|$ piecewise:$|x - 5| = \begin{cases} -(x - 5), & 2 \le x < 5 \\ x - 5, & 5 \le x \le 8 \end{cases}$.Thus the integral becomes:$\int_{2}^{8} |x - 5| \, dx = \int_{2}^{5} -(x - 5) \, dx + \int_{5}^{8} (x - 5) \, dx$.Simplify the first integrand: $-(x - 5) = 5 - x$.So $\int_{2}^{5} (5 - x) \, dx + \int_{5}^{8} (x - 5) \, dx$.Evaluate each integral using basic antiderivatives:$\int (5 - x) \, dx = 5x - \frac{x^2}{2}$.$\int (x - 5) \, dx = \frac{x^2}{2} - 5x$.Apply limits:First part: $\left[5x - \frac{x^2}{2}\right]_{2}^{5} = \left(25 - \frac{25}{2}\right) - \left(10 - 2\right) = \frac{25}{2} - 8$.Second part: $\left[\frac{x^2}{2} - 5x\right]_{5}^{8} = \left(32 - 40\right) - \left(\frac{25}{2} - 25\right) = -8 + \frac{25}{2}$.Add the two results: $\left(\frac{25}{2} - 8\right) + \left(-8 + \frac{25}{2}\right) = \frac{25}{2} + \frac{25}{2} - 16 = 25 - 16 = 9$.Answer:$9$ (Option D).