Question:

MATCH THE FOLLOWING: Column 1: (A) $f(x) = \begin{cases} x\left(\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right] + \cdots + \left[\frac{8}{x}\right]\right), & x \neq 0 \\ 9k, & x = 0 \end{cases}$ The value of $k$ such that $f$ is continuous at $x = 0$ is (B) $f(x) = \begin{cases} \left(1 + xe^{-1/x^2}\sin\frac{1}{x^3}\right)^{1/x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ The value of $k$ such that $f$ is continuous at $x = 0$ is (C) $f: [0, \infty) \to \mathbb{R}; f(x) = \begin{cases} \left(2\sin\sqrt{x} + \sqrt{x}\sin\frac{1}{x}\right)^x, & x > 0 \\ k, & x = 0 \end{cases}$ The value of $k$ such that $f$ is continuous at $x = 0$ is (D) $f: (0, \pi) \to \mathbb{R}; f(x) = \begin{cases} \left(\frac{1 - \sin x}{(\pi - 2x)^2} \cdot \ln(1 + x^2 - 4\pi x + 4x^2)\right)^{\tan x}, & x \neq \frac{\pi}{2} \\ k, & x = \frac{\pi}{2} \end{cases}$ The value of $8\sqrt[4]{k}$ such that $f$ is continuous at $x = \frac{\pi}{2}$ is Column 2: (p) 1 (q) 2 (r) 3 (s) 4