CUET PG MCA Continuity Previous Year Questions (PYQs)

$f(x)=\tan^2 x$ is not defined at $x=\pi/2$, hence it is not continuous there.

So Assertion A is false.

$g(x)=x^2$ is a polynomial, hence continuous everywhere, including at $x=\pi/2$.

So Reason R is true.

We have: \[ f(x) = \begin{cases} x \sin\!\left(\tfrac{1}{x}\right), & x \neq 0, \\[6pt] 0, & x = 0. \end{cases} \]

Step 1: Continuity at \(x=0\)

\[ \lim_{x \to 0} x \sin\!\left(\tfrac{1}{x}\right). \] Since \(|\sin(1/x)| \leq 1\), \[ -|x| \;\leq\; x \sin\!\left(\tfrac{1}{x}\right) \;\leq\; |x|. \] By the squeeze theorem, \[ \lim_{x \to 0} f(x) = 0 = f(0). \] ✅ Thus, \(f(x)\) is continuous everywhere.