Question

If a twice differentiable function satisfying a relation $f\left(x^2 y\right)=x^2 f(y)+y f\left(x^2\right) \forall x, y>0$ and $f^{\prime}(1)=1$ then the value of $f^{\prime \prime}\left(\frac{1}{7}\right)$ is

• A. 7
• B. $\frac{7}{2}$
• C. $\frac{7}{3}$
• D. $\frac{7}{4}$

Answer 1

Answer: (A)

$ f \left( x ^2 y \right)= x ^2 f ( y )+ yf \left( x ^2\right) \forall x , y >0 $
differentiate w.r.t. $x$ keeping y constant
$2 x y f^{\prime}\left(x^2 y\right)=f(y) \cdot 2 x+y \cdot 2 x f^{\prime}\left(x^2\right) $
$\therefore y f^{\prime}\left(x^2 y\right)=f(y)+y f^{\prime}\left(x^2\right)$
Put $x =1$
$y f^{\prime}(y)=f(y)+y f^{\prime}(1)=f(y)+y$.....(1)
Now again differentiate w.r.t. $y$
$y f^{\prime \prime}(y)+f^{\prime}(y)=f^{\prime}(y)+1 $
$\therefore f^{\prime \prime}(y)=\frac{1}{y} \Rightarrow f^{\prime \prime}\left(\frac{1}{7}\right)=7$