UPSC NDA I 2016 Mathematics Solved Paper

Concept:The absolute value function $|x|$ is defined piecewise: $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$.We rewrite $f(x) = x|x| - 1$ in piecewise form, then equate to $g(x)$ piecewise to find intersection points.Explanation:Step 1: Write $f(x)$ piecewise:$f(x) = \begin{cases} x^2 - 1, & x \ge 0 \\ -x^2 - 1, & x < 0 \end{cases}$.Step 2: For $x > 0$: set $f(x) = g(x)$ where $g(x) = \frac{3x}{2}$.$x^2 - 1 = \frac{3x}{2} \Rightarrow 2x^2 - 3x - 2 = 0$.Factor: $(2x+1)(x-2)=0 \Rightarrow x = 2$ (since $x > 0$).Then $y = f(2) = 2^2 - 1 = 3$. So point is $(2,3)$.Step 3: For $x < 0$: $f(x) = -x^2 - 1$, $g(x) = 2x$.Set equal: $-x^2 - 1 = 2x \Rightarrow x^2 + 2x + 1 = 0 \Rightarrow (x+1)^2 = 0 \Rightarrow x = -1$.Then $y = f(-1) = -(-1)^2 - 1 = -2$. So point is $(-1,-2)$.Step 4: Both points satisfy the respective conditions, so they are valid intersection points.Answer:The curves intersect at $(2,3)$ and $(-1,-2)$. Hence option C is correct.