UPSC NDA 2 2016 Mathematics Solved Paper

Concept:A function is continuous at $x=a$ if left-hand limit equals right-hand limit equals $f(a)$. It is differentiable at $x=a$ if left-hand derivative equals right-hand derivative. The greatest value in a closed interval occurs at a critical point or an endpoint.Explanation:Given:$$f(x)= \begin{cases} 3x^{2}+12x-1, & -1\le x\le 2 \\ 37-x, & 2 < x \le 3 \end{cases}$$Statement 1 – Continuity at $x=2$:Left-hand limit: $\displaystyle\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(3x^{2}+12x-1)=3(4)+24-1=35$.Right-hand limit: $\displaystyle\lim_{x\to2^{+}}f(x)=\lim_{x\to2^{+}}(37-x)=37-2=35$.Also $f(2)=3(4)+24-1=35$.Thus $\text{LHL}=\text{RHL}=f(2)$, so $f$ is continuous at $x=2$.Statement 1 is correct.Statement 2 – Greatest value at $x=2$:Evaluate $f$ at the endpoints of the piecewise intervals and at $x=2$:$f(-1)=3-12-1=-10$,$f(2)=35$,$f(3)=37-3=34$.Since $35$ is the largest among these, $f$ attains its greatest value at $x=2$.Statement 2 is correct.Statement 3 – Differentiability at $x=2$:Derivative from the left: $f'(x)=6x+12$ for $x\le2$, so $\text{LHD}=6(2)+12=24$.Derivative from the right: $f'(x)=-1$ for $x>2$, so $\text{RHD}=-1$.Since $24 \ne -1$, the left-hand derivative does not equal the right-hand derivative.Hence $f$ is not differentiable at $x=2$.Statement 3 is incorrect.Answer:Only statements 1 and 2 are correct. Therefore the correct code is A (1 and 2 only).