UPSC NDA II 2024 Mathematics Paper

Concept:The absolute value function splits the integral at points where the expression inside changes sign.We find the roots of $x^2 - x - 2$ to determine where it is positive or negative.Explanation:Step 1: Factor the quadratic: $x^2 - x - 2 = (x-2)(x+1)$. Roots at $x = -1$ and $x = 2$.Step 2: On the interval $[1,3]$, the sign changes at $x=2$. For $1 \le x < 2$, the expression is negative, so $|x^2-x-2| = -(x^2-x-2)$. For $2 \le x \le 3$, it is positive, so $|x^2-x-2| = x^2-x-2$.Step 3: Write the integral as sum of two parts:$\int_{1}^{3} f(x)\,dx = \int_{1}^{2} [-(x^2-x-2)]\,dx + \int_{2}^{3} (x^2-x-2)\,dx$.Step 4: Compute the antiderivative $F(x) = \frac{x^3}{3} - \frac{x^2}{2} - 2x$.First part: $\int_{1}^{2} -(x^2-x-2)\,dx = -[F(2)-F(1)]$.$F(2) = \frac{8}{3} - 2 - 4 = \frac{8}{3} - 6 = -\frac{10}{3}$.$F(1) = \frac{1}{3} - \frac{1}{2} - 2 = \frac{2-3-12}{6} = -\frac{13}{6}$.So $- \left( -\frac{10}{3} + \frac{13}{6} \right) = - \left( -\frac{20}{6} + \frac{13}{6} \right) = - \left( -\frac{7}{6} \right) = \frac{7}{6}$.Step 5: Second part: $\int_{2}^{3} (x^2-x-2)\,dx = F(3)-F(2)$.$F(3) = \frac{27}{3} - \frac{9}{2} - 6 = 9 - 4.5 - 6 = -1.5 = -\frac{3}{2}$.$F(3)-F(2) = -\frac{3}{2} - (-\frac{10}{3}) = -\frac{3}{2} + \frac{10}{3} = \frac{-9+20}{6} = \frac{11}{6}$.Step 6: Total integral = $\frac{7}{6} + \frac{11}{6} = \frac{18}{6} = 3$.Answer:$3$ (Option B)