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Derivative of f(x)=∫dt/(1+|x−t|), find f′(1/2) | JEE

JEE Maths question with a full step-by-step solution.

Question

f(x)=\displaystyle\int_{0}^{1}\dfrac{dt}{1+|x-t|}

, then the value of

f'\!\left(\dfrac12\right)

0

correct

1

2

3

Solution

Step 1: Remove the absolute value by splitting the range of

t

t=x

(valid for

0<x<1

). For

t<x

|x-t|=x-t

; for

t>x

|x-t|=t-x

. Thus

f(x)=\int_{0}^{x}\dfrac{dt}{1+x-t}+\int_{x}^{1}\dfrac{dt}{1+t-x}.

Step 2: Integrate each piece. For the first, with the linear argument

1+x-t

\int_{0}^{x}\dfrac{dt}{1+x-t}=\Big[-\ln(1+x-t)\Big]_{0}^{x}=-\ln 1+\ln(1+x)=\ln(1+x).

For the second,

\int_{x}^{1}\dfrac{dt}{1+t-x}=\Big[\ln(1+t-x)\Big]_{x}^{1}=\ln(2-x)-\ln 1=\ln(2-x).

Step 3: Hence

f(x)=\ln(1+x)+\ln(2-x)

, valid on

(0,1)

. Differentiate:

f'(x)=\dfrac{1}{1+x}-\dfrac{1}{2-x}.

Step 4: Substitute

x=\dfrac12

f'\!\left(\dfrac12\right)=\dfrac{1}{3/2}-\dfrac{1}{3/2}=\dfrac{2}{3}-\dfrac{2}{3}=0.

Correct answer: (1)

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