UPSC NDA Math Model Paper 1

The given function is $f(x)=\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$ for $x < 0$ and $f(x)=\frac{1+2x}{1-x}$ for $x>0.$ Let's calculate the left-hand $(x \rightarrow 0^{-})$ and the right-hand $(x \rightarrow 0^{+})$ limits of the function: $\lim\limits_{x \rightarrow 0^{-}} f(x) = \lim\limits_{x \rightarrow 0} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$ which is an indeterminate from $\frac{0}{0}$ . Let's rationalize by multiplying with the conjugate:$= \lim\limits_{x \rightarrow 0} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$ $\times \frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}$ $= \lim\limits_{x \rightarrow 0} \frac{1}{x} \times \frac{(1+kx)-(1-kx)}{\sqrt{1+kx}+\sqrt{1-kx}}$ $= \lim\limits_{x \rightarrow 0} \frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}$ $=k$ And, $\lim\limits_{x \rightarrow 0^{+}} f(x) = \lim\limits_{x \rightarrow 0} \frac{1+2x}{1-x} = 1$ Since, f(x) is given to be continuous at x = 0, both the left-hand and the right-hand side limits must be equal. ∴ k = 1.