UPSC NDA I 2016 Mathematics Solved Paper

Concept:A function is continuous at a point if left-hand limit, right-hand limit, and the function value are equal.A function is differentiable at a point if left-hand derivative equals right-hand derivative.Explanation:Given: $f(x) = \left|x-1\right| + x^2$, for $x \in \mathbb{R}$.Write $f(x)$ as a piecewise function:$$f(x)=\begin{cases} x^2 - x + 1, & x < 1 \\ x^2 + x - 1, & x \ge 1 \end{cases}$$Check continuity at $x=1$:Left-hand limit: $\displaystyle\lim\limits_{x \to 1^{-}} f(x) = 1^2 - 1 + 1 = 1$Right-hand limit: $\displaystyle\lim\limits_{x \to 1^{+}} f(x) = 1^2 + 1 - 1 = 1$Function value: $f(1)=1^2+1-1=1$Since $LHL = RHL = f(1)$, $f$ is continuous at $x=1$.Now check differentiability at $x=1$:Differentiate each piece: $f'(x)=\begin{cases} 2x-1, & x<1 \\ 2x+1, & x>1 \end{cases}$Left-hand derivative: $\displaystyle\lim\limits_{x \to 1^{-}} f'(x) = 2(1)-1 = 1$Right-hand derivative: $\displaystyle\lim\limits_{x \to 1^{+}} f'(x) = 2(1)+1 = 3$Since $LHD \neq RHD$, $f$ is not differentiable at $x=1$.Answer:Option B: $f(x)$ is continuous but not differentiable at $x=1$.