The function f: R→ R defined by f(x) =≤ft […] . is

The correct answer is Option (1) → Continuous but not differentiable at x = 1

Given:

$f(x) = \begin{cases} x^2, & x \geq 1 \\ x, & x < 1 \end{cases}$

Step 1: Check continuity at $x = 1$

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1$

$f(1) = 1$

Since LHL = RHL = f(1), function is continuous at $x = 1$

Step 2: Check differentiability at $x = 1$

$f'(x) = \begin{cases} 2x, & x > 1 \\ 1, & x < 1 \end{cases}$

Left-hand derivative at $x = 1$: $f'_-(1) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = 1$

Right-hand derivative at $x = 1$: $f'_+(1) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^+} (x + 1) = 2$

Since LHD ≠ RHD, function is not differentiable at $x = 1$

Final Answer: Continuous but not differentiable at $x = 1$