CUET UG 2025 — Mathematics Calculus

For the function to be continuous at $x = 0$x=0, the left-hand limit, right-hand limit, and the value at $x = 0$x=0 must all be equal.

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Since $x = 0$x=0 satisfies $x \leq 0$x≤0, the first piece of the function applies:

$$f(0) = \alpha(0^2 - 2(0) + 1)$$f(0)=α(02−2(0)+1)

$$f(0) = \alpha(1)$$f(0)=α(1)

$$f(0) = \alpha$$f(0)=α

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For $x > 0$x>0, the function is $f(x) = 2x + 1$f(x)=2x+1.

The right-hand limit as $x$x approaches $0$0:

$$\lim_{x \to 0^+} f(x) = 2(0) + 1$$x→0+lim​f(x)=2(0)+1

$$\lim_{x \to 0^+} f(x) = 1$$x→0+lim​f(x)=1

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For $x < 0$x<0, the function is $f(x) = \alpha(x^2 - 2x + 1)$f(x)=α(x2−2x+1).

The left-hand limit as $x$x approaches $0$0:

$$\lim_{x \to 0^-} f(x) = \alpha(0^2 - 2(0) + 1)$$x→0−lim​f(x)=α(02−2(0)+1)

$$\lim_{x \to 0^-} f(x) = \alpha$$x→0−lim​f(x)=α

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For continuity at $x = 0$x=0:

$$f(0) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x)$$f(0)=x→0+lim​f(x)=x→0−lim​f(x)

$$\alpha = 1 = \alpha$$α=1=α

Therefore, $\alpha = 1$α=1