f(x) = [x] , where [ ] represents greatest integer function. (A) For…

The correct answer is Option (4) → (B), (E) only

The greatest integer function [x] is the largest integer not greater than x. Near x = 2:

    \(f(x) = 2\) for \(2 \le x \lt 3\)
    \(f(x) = 1\) for \(1 \le x \lt 2\)
    \(f(2) = 2\)

Right hand derivative at x = 2 (h > 0, small):

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

Left hand derivative at x = 2 (h < 0, small):

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

This limit does not exist (it grows without bound). So the right hand derivative is 0, the left hand derivative does not exist, and f is not differentiable at x = 2.

Checking the statements

(A) For \(2 \le x \lt 3\), \([x] = 3\) — Incorrect. On this interval [x] = 2.

(B) For \(2 \le x \lt 3\), \([x] = 2\) — Correct. 2 is the largest integer not greater than x here.

(C) Right hand derivative of f at x = 2 is not defined — Incorrect. f is constant 2 just to the right of 2, so the right hand derivative is 0.

(D) Left hand derivative of f at x = 2 is zero — Incorrect. The quotient is −1/h, which has no finite limit as \(h \to 0^-\).

(E) f(x) is not differentiable at x = 2 — Correct. f jumps from 1 to 2 at x = 2, so it is not even continuous there.