Select the correct answer from the options (A), (B), (C) and (D) as given below.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Assertion (A): Consider the function defined as f(x)=∣x∣+∣x−1∣, x∈R. Then f(x) is not differentiable at x=0 and x=1.
Reason (R): Suppose f be defined and continuous on (a,b) and c∈(a,b), then f(x) is not differentiable at x=c if
h→0−limhf(c+h)−f(c)=h→0+limhf(c+h)−f(c).
Exam answer
f(x)=⎩⎨⎧1−2x,1,2x−1,x<00≤x<1x≥1Why?
Each modulus changes form where its inside is 0, at x=0 and x=1. Remove the bars on each piece:
x<00≤x<1x≥1:−x−(x−1):x−(x−1):x+(x−1)
(For x<0 both insides are negative, so ∣x∣=−x and ∣x−1∣=−(x−1). For 0≤x<1 only x−1 is negative. For x≥1 neither is.)
−x−(x−1)x−(x−1)x+(x−1)=−x−x+1=1−2x=x−x+1=1=2x−1
These simplify to 1−2x,1 and 2x−1. So f is continuous everywhere, and f(0)=f(1)=1.
f(0)f(1)=1(middle piece)=2(1)−1=1
At x=0, with f(0)=1:
LHDRHD=h→0−limh(1−2h)−1=−2=h→0+limh1−1=0
At x=1, with f(1)=1:
LHDRHD=h→0−limh1−1=0=h→0+limh(1+2h)−1=2Why?
A one-sided derivative at c is the limit, as h tends to 0 from one side, of
hf(c+h)−f(c)
Use the piece of f on that side.
At 0: for small h<0,f(h)=1−2h; for small h>0,f(h)=1.
At 1: for small h<0,f(1+h)=1; for small h>0,f(1+h)=2(1+h)−1=1+2h.
Simplify each top, then cancel h:
At 0, leftAt 0, rightAt 1, leftAt 1, right:h(1−2h)−1=h−2h=−2:h1−1=h0=0:h1−1=h0=0:h(1+2h)−1=h2h=2
At x=0: LHD =−2, RHD =0. At x=1: LHD =0, RHD =2.
LHD = RHD at both points, so f is not differentiable at x=0 and x=1. Assertion (A) is true.
Reason (R) is true: a function is differentiable at c only if LHD = RHD there. It is the rule used above, so it explains the Assertion.
Why?
Differentiable at c means that the difference quotient used above has a limit as h→0. A limit exists only when the left and right limits are equal, so unequal one-sided derivatives mean no derivative.
The Assertion was proved with exactly this rule, so the Reason explains it: option (A).
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Where this comes from
Chapter 5, Continuity and Differentiability: Differentiability of |x| + |x − 1|.
Type 3. Is f differentiable at a point? Compare LHD and RHD
Spot it: a modulus such as ∣x−1∣, or a function in pieces, and the word "differentiable".
Here: Write ∣x∣+∣x−1∣ in pieces: 1−2x, 1, 2x−1. At 0, LHD is −2 and RHD is 0; at 1, LHD is 0 and RHD is 2. Unequal at both, so A is true and R is the reason: option (A).