Continuity & DifferentiabilitymediumFree

Continuity & Differentiability: Twice Differentiable Function Satisfies Relation Value

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

Question

If a twice differentiable function satisfies the relation

f\left(x^2y\right) = x^2f(y)+y\,f\left(x^2\right) \qquad \forall\ x,y>0

and

f(1) = 1

, then the value of

f''\left(\dfrac17\right)

\dfrac17

7

correct

3

\dfrac27

Solution

Step 1: Differentiating the relation with respect to

x

, with

y

fixed,

\frac{d}{dx}f\left(x^2y\right) = f'\left(x^2y\right)\cdot2xy

\frac{d}{dx}\left[x^2f(y)+y\,f\left(x^2\right)\right] = 2x\,f(y)+2xy\,f'\left(x^2\right)

2xy\,f'\left(x^2y\right) = 2x\,f(y)+2xy\,f'\left(x^2\right)

Step 2:

x>0

, so divide by

2x

and put

x = 1

y\,f'\left(x^2y\right) = f(y)+y\,f'\left(x^2\right) \quad\xrightarrow{\ x = 1\ }\quad y\,f'(y) = f(y)+y\,f'(1)

Step 3: Differentiating in

y

y\,f''(y)+f'(y) = f'(y)+f'(1) \quad\Longrightarrow\quad y\,f''(y) = f'(1) \quad\Longrightarrow\quad f''(y) = \frac{f'(1)}{y}

which is valid for

y>0

. Step 4: The condition intended by the question is

f'(1) = 1

(see the note below), so

f''(y) = \frac1y

Putting

y = 1

in the given relation gives

f\left(x^2\right) = x^2f(1)+f\left(x^2\right)

, i.e.

f(1) = 0

. Integrating twice with

f'(1) = 1

and

f(1) = 0

gives

f'(y) = \ln y+1

and

f(y) = y\ln y

, and

f(t) = t\ln t

does satisfy the functional relation:

f\left(x^2y\right) = x^2y\ln\left(x^2y\right) = x^2y\ln y+x^2y\ln x^2 = x^2f(y)+y\,f\left(x^2\right).

Step 5:

f''\left(\frac17\right) = \frac1{1/7} = 7

Answer: (2)

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