UPSC NDA II 2023 Mathematics Solved Paper

Concept:For a function to be differentiable at a point, it must be continuous there, and its left‑hand derivative must equal its right‑hand derivative.Explanation:The function is defined as:$f(x)=\begin{cases} ax(x+1)+b, & x<1 \\ x-1, & 1\le x\le 2 \end{cases}$First, find the derivative for each piece:$f'(x)=\begin{cases} 2ax+a, & x<1 \\ 1, & 1\le x\le 2 \end{cases}$Differentiability at $x=1$ requires left‑hand derivative = right‑hand derivative.Left‑hand derivative at $x=1$: $2a(1)+a = 3a$.Right‑hand derivative at $x=1$: $1$.Thus $3a = 1 \implies a = \frac{1}{3}$.Since the function is differentiable at $x=1$, it is also continuous there.Continuity requires left‑hand limit = right‑hand limit = $f(1)$.Left‑hand limit at $x=1$: $a(1)(1+1)+b = 2a+b$.Right‑hand limit at $x=1$: $1-1 = 0$.Thus $2a+b = 0$.Substitute $a = \frac{1}{3}$: $2\cdot\frac{1}{3}+b = 0 \implies b = -\frac{2}{3}$.Hence $a+b = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.Answer:$-\frac{1}{3}$