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Exercise 10.1 · Q3

Q.Determine whether the following function is differentiable at the indicated values.

(i) f(x)=x ∣x∣f(x) = x\,|x| at x=0x = 0
(ii) f(x)=∣x2−1∣f(x) = |x^2 - 1| at x=1x = 1
(iii) f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1| at x=0,1x = 0, 1
(iv) f(x)=sin⁡∣x∣f(x) = \sin|x| at x=0x = 0
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✓ Free question

Step 1. (i) f(x)=x∣x∣f(x)=x|x|: f(x)=x2f(x)=x^2 for x≥0x\ge0, f(x)=−x2f(x)=-x^2 for x<0x<0; f(0)=0f(0)=0.

LHD: for h→0−h\to0^-, f(h)=−h2f(h)=-h^2, quotient =−h2−0h=−h→0=\dfrac{-h^2-0}{h}=-h\to0.

RHD: for h→0+h\to0^+, f(h)=h2f(h)=h^2, quotient =h2−0h=h→0=\dfrac{h^2-0}{h}=h\to0.

Both one-sided derivatives equal 00, so ff IS differentiable at x=0x=0, with f′(0)=0f'(0)=0.

Step 2. (ii) f(x)=∣x2−1∣f(x)=|x^2-1|. Near x=1x=1: for x∈(−1,1)x\in(-1,1), x2−1<0x^2-1<0 so f(x)=1−x2f(x)=1-x^2; for x>1x>1, x2−1>0x^2-1>0 so f(x)=x2−1f(x)=x^2-1. f(1)=0f(1)=0.

LHD: h→0−h\to0^-, x=1+h<1x=1+h<1, f(1+h)=1−(1+h)2=−2h−h2f(1+h)=1-(1+h)^2=-2h-h^2. Quotient =−2h−h2h=−2−h→−2=\dfrac{-2h-h^2}{h}=-2-h\to-2.

RHD: h→0+h\to0^+, x=1+h>1x=1+h>1, f(1+h)=(1+h)2−1=2h+h2f(1+h)=(1+h)^2-1=2h+h^2. Quotient =2h+h2h=2+h→2=\dfrac{2h+h^2}{h}=2+h\to2.

LHD =−2≠2==-2\neq2= RHD, so ff is NOT differentiable at x=1x=1.

Step 3. (iii) f(x)=∣x∣+∣x−1∣f(x)=|x|+|x-1|.

At x=0x=0: f(0)=0+1=1f(0)=0+1=1. For h<0h<0 (small), f(h)=−h+(1−h)=1−2hf(h)=-h+(1-h)=1-2h [since h<0h<0 and h<1h<1]. LHD =(1−2h)−1h=−2hh=−2=\dfrac{(1-2h)-1}{h}=\dfrac{-2h}{h}=-2.

For h>0h>0 (small, <1<1), f(h)=h+(1−h)=1f(h)=h+(1-h)=1. RHD =1−1h=0=\dfrac{1-1}{h}=0.

LHD =−2≠0==-2\neq0= RHD: NOT differentiable at x=0x=0.

At x=1x=1: f(1)=1+0=1f(1)=1+0=1. For h<0h<0 small, x=1+h∈(0,1)x=1+h\in(0,1): f(1+h)=(1+h)+(1−(1+h))=(1+h)+(−h)=1f(1+h)=(1+h)+(1-(1+h))=(1+h)+(-h)=1. LHD =1−1h=0=\dfrac{1-1}{h}=0.

For h>0h>0 small, x=1+h>1x=1+h>1: f(1+h)=(1+h)+h=1+2hf(1+h)=(1+h)+h=1+2h. RHD =2hh=2=\dfrac{2h}{h}=2.

LHD =0≠2==0\neq2= RHD: NOT differentiable at x=1x=1.

Step 4. (iv) f(x)=sin⁡∣x∣f(x)=\sin|x|; f(0)=0f(0)=0.

RHD: h→0+h\to0^+, ∣h∣=h|h|=h, f(h)=sin⁡hf(h)=\sin h. Quotient =sin⁡hh→1=\dfrac{\sin h}{h}\to1 (standard limit). RHD=1=1.

LHD: h→0−h\to0^-, ∣h∣=−h|h|=-h, f(h)=sin⁡(−h)f(h)=\sin(-h). Quotient =sin⁡(−h)h=−sin⁡hh→−1=\dfrac{\sin(-h)}{h}=\dfrac{-\sin h}{h}\to-1 (since sin⁡h/h→1\sin h/h\to1 regardless of which side hh approaches 00 from). LHD=−1=-1.

LHD =−1≠1==-1\neq1= RHD: even though sin⁡\sin is smooth, the outer ∣x∣|x| still produces a genuine corner at x=0x=0 (near 00, sin⁡∣x∣≈∣x∣\sin|x|\approx|x|), so ff is NOT differentiable at x=0x=0.

✓Final answer

(i) Differentiable at x=0x=0, f′(0)=0f'(0)=0.

(ii) Not differentiable at x=1x=1.

(iii) Not differentiable at x=0x=0 nor at x=1x=1.

(iv) Not differentiable at x=0x=0.

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