MHTCET 23 Apr 2025 Shift 2 Paper

Concept:A function is continuous at a point if left-hand limit, right-hand limit and the value of the function at that point are all equal.Explanation:Rewrite the function by removing the absolute value.Since $|x - x^2| = x - x^2$ when $x - x^2 \ge 0$, i.e. for $0 \le x \le 1$,$$f(x) = 2x - (x - x^2) = x + x^2.$$For $x - x^2 < 0$, i.e. for $x < 0$ or $x > 1$,$$f(x) = 2x - (x^2 - x) = 3x - x^2.$$Check continuity at $x = 1$:LHL $= \lim_{x \to 1^-} f(x) = \lim_{x \to 1} (x + x^2) = 2$.RHL $= \lim_{x \to 1^+} f(x) = \lim_{x \to 1} (3x - x^2) = 2$.Also, $f(1) = 1 + 1^2 = 2$.Since LHL = RHL = f(1), the function is continuous at $x = 1$.Check continuity at $x = 0$:LHL $= \lim_{x \to 0^-} f(x) = \lim_{x \to 0} (3x - x^2) = 0$.RHL $= \lim_{x \to 0^+} f(x) = \lim_{x \to 0} (x + x^2) = 0$.Also, $f(0) = 0$.Hence, the function is continuous at $x = 0$ as well.Answer:The function is continuous at $x = 1$, so option A is correct.