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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(x2y)=x2f(y)+y f(x2)∀ x,y>0f\left(x^2y\right) = x^2f(y)+y\,f\left(x^2\right) \qquad \forall\ x,y>0
 and f(1)=1f(1) = 1, then the value of f′′(17)f''\left(\dfrac17\right) is
A17\dfrac17
B77correct
C33
D27\dfrac27
Solution
Step 1: Differentiating the relation with respect to xx, with yy fixed,
ddxf(x2y)=f′(x2y)⋅2xy\frac{d}{dx}f\left(x^2y\right) = f'\left(x^2y\right)\cdot2xy
ddx[x2f(y)+y f(x2)]=2x f(y)+2xy f′(x2)\frac{d}{dx}\left[x^2f(y)+y\,f\left(x^2\right)\right] = 2x\,f(y)+2xy\,f'\left(x^2\right)
 so
2xy f′(x2y)=2x f(y)+2xy f′(x2)2xy\,f'\left(x^2y\right) = 2x\,f(y)+2xy\,f'\left(x^2\right)
 Step 2: x>0x>0, so divide by 2x2x and put x=1x = 1:
y f′(x2y)=f(y)+y f′(x2)→ x=1 y f′(y)=f(y)+y f′(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 yy,
y f′′(y)+f′(y)=f′(y)+f′(1)⟹y f′′(y)=f′(1)⟹f′′(y)=f′(1)yy\,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>0y>0. Step 4: The condition intended by the question is f′(1)=1f'(1) = 1 (see the note below), so
f′′(y)=1yf''(y) = \frac1y
 Putting y=1y = 1 in the given relation gives f(x2)=x2f(1)+f(x2)f\left(x^2\right) = x^2f(1)+f\left(x^2\right), i.e. f(1)=0f(1) = 0. Integrating twice with f′(1)=1f'(1) = 1 and f(1)=0f(1) = 0 gives f′(y)=ln⁡y+1f'(y) = \ln y+1 and f(y)=yln⁡yf(y) = y\ln y, and f(t)=tln⁡tf(t) = t\ln t does satisfy the functional relation:
f(x2y)=x2yln⁡(x2y)=x2yln⁡y+x2yln⁡x2=x2f(y)+y f(x2).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′′(17)=11/7=7f''\left(\frac17\right) = \frac1{1/7} = 7
 Answer: (2)
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