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NCERT Exemplar · Q20

Q.Examine the differentiability of ff, where ff is defined by f(x)={x[x],0≤x<2(x−1)x,2≤x<3f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x - 1)x, & 2 \le x < 3 \end{cases} at x=2x = 2 (here [x][x] denotes the greatest integer function).

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Checking one-sided derivatives at x=2x=2: f−′(2)=1f'_-(2)=1 and f+′(2)=3f'_+(2)=3. Since these disagree, ff is not differentiable at x=2x=2 (even though it is continuous there).

Setting Up

The formula for ff changes at x=2x=2, so that is the only point to examine. For a piecewise function, always check continuity first, then compare the one-sided derivatives.

For 1≤x<21 \le x < 2, [x]=1[x]=1, so the first piece is f(x)=x[x]=xf(x)=x[x]=x. For 2≤x<32 \le x < 3, f(x)=(x−1)xf(x)=(x-1)x, so f(2)=(2−1)(2)=2f(2)=(2-1)(2)=2.

Step 1 — Continuity at x=2x=2

lim⁡x→2−f(x)=lim⁡x→2−x=2,lim⁡x→2+f(x)=lim⁡x→2+(x−1)x=(1)(2)=2.\lim_{x\to 2^-}f(x)=\lim_{x\to2^-}x=2,\qquad \lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x-1)x=(1)(2)=2.

Both one-sided limits equal f(2)=2f(2)=2, so ff is continuous at x=2x=2 — differentiability is still possible.

Step 2 — Left-hand derivative

For h<0h<0 small, 2+h∈(1,2)2+h\in(1,2), so f(2+h)=2+hf(2+h)=2+h:

f−′(2)=lim⁡h→0−(2+h)−2h=lim⁡h→0−hh=1.f'_-(2)=\lim_{h\to0^-}\frac{(2+h)-2}{h}=\lim_{h\to0^-}\frac{h}{h}=1.

Step 3 — Right-hand derivative

For h>0h>0 small, 2+h∈(2,3)2+h\in(2,3), so f(2+h)=(1+h)(2+h)=2+3h+h2f(2+h)=(1+h)(2+h)=2+3h+h^2: …

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