CUET UG 2025 — Mathematics Calculus

The function is defined as:

$$f(x) = \begin{cases} x^2, & x \ge 1 \\ x, & x < 1 \end{cases}$$f(x)={x2,x,​x≥1x<1​

The function switches formulas at $x = 1$x=1, so continuity and differentiability need to be checked at this point.

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For continuity at $x = 1$x=1, the left-hand limit, right-hand limit, and function value must all be equal.

$$f(1) = 1^2 = 1$$f(1)=12=1

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

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$$x→1+lim​f(x)=x→1+lim​x2=12=1

Since all three values equal $1$1, the function is continuous at $x = 1$x=1.

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For differentiability at $x = 1$x=1, the left-hand derivative and right-hand derivative must be equal.

For $x < 1$x<1, $f(x) = x$f(x)=x

$$f'(x) = 1$$f′(x)=1

$$f'(1^-) = 1$$f′(1−)=1

For $x \ge 1$x≥1, $f(x) = x^2$f(x)=x2

$$f'(x) = 2x$$f′(x)=2x

$$f'(1^+) = 2(1) = 2$$f′(1+)=2(1)=2

Since $1 \neq 2$1=2, the left-hand derivative does not equal the right-hand derivative. The function is not differentiable at $x = 1$x=1.

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The function is continuous but not differentiable at $x = 1$x=1.