UPSC NDA 2 2019 Mathematics Solved Paper

Concept:For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function's value at that point must all be equal.Explanation:The given function is $f(x) = \begin{cases} 2x + \frac{1}{4}, & x < 0 \\ k, & x = 0 \\ (x + \frac{1}{2})^2, & x > 0 \end{cases}$.We need to find $k$ such that $f$ is continuous at $x=0$.Compute the left-hand limit as $x$ approaches $0$ from the left:$\lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \left[2(-h) + \frac{1}{4}\right] = \frac{1}{4}$.Compute the right-hand limit as $x$ approaches $0$ from the right:$\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} (h + \frac{1}{2})^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.Since both one-sided limits are $\frac{1}{4}$, the overall limit exists and equals $\frac{1}{4}$.For continuity at $x=0$, we require $\lim_{x \to 0} f(x) = f(0)$.Given $f(0) = k$, we get $k = \frac{1}{4}$.Answer:The value of $k$ is $\frac{1}{4}$, which corresponds to option A.