MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Continuity of a function is checked by comparing its left-hand limit, right-hand limit, and the value at the point.Explanation:For $x \neq 0$, we write $f(x)=\frac{x-|x|}{x}$ by splitting according to the sign of $x$.When $x > 0$, we have $|x| = x$, so $f(x)=\frac{x-x}{x}=0$.When $x < 0$, we have $|x| = -x$, so $f(x)=\frac{x-(-x)}{x}=\frac{2x}{x}=2$.Also, it is given that $f(0)=2$.Thus, $f(x) = \begin{cases} 2, & x < 0 \\ 2, & x = 0 \\ 0, & x > 0 \end{cases}$.For $x < 0$, $f(x)$ is constant, so it is continuous there.For $x > 0$, $f(x)$ is constant, so it is continuous there.At $x=0$, the left-hand limit is $\lim_{x \to 0^{-}} f(x) = 2$.The right-hand limit is $\lim_{x \to 0^{+}} f(x) = 0$.Since the two one-sided limits are unequal, $\lim_{x \to 0} f(x)$ does not exist.Therefore, $f$ is not continuous at $x=0$, but it is continuous for all other real numbers.Answer:$f$ is continuous for all $x$ except at $x=0$.Hence, the correct option is B.