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

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

Continuity at x = 1:

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

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

    \(f(1) = 1^2 = 1\)

All three are equal, so f is continuous at x = 1.

Differentiability at x = 1: compare the left and right derivatives.

    \(\displaystyle Lf'(1) = \lim_{h \to 0^-} \frac{(1 + h) - 1}{h} = 1\)

    \(\displaystyle Rf'(1) = \lim_{h \to 0^+} \frac{(1 + h)^2 - 1}{h} = \lim_{h \to 0^+} (2 + h) = 2\)

Since 1 ≠ 2, f is not differentiable at x = 1.

Option (4) does not hold: near x = 0, f(x) = x, which is differentiable there.