Question

If a twice differentiable function satisfying a relation $f\left(x^{2}y\right)=x^{2}f(y)+yf\left(x^2\right)\forall x,y>0$ 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
$2xy{f}^{\prime }\left(x^{2}y\right)=f(y)\cdot 2x+y\cdot 2x{f}^{\prime }\left(x^2\right)$
$\therefore y{f}^{\prime }\left(x^{2}y\right)=f(y)+y{f}^{\prime }\left(x^2\right)$
Put $x=1$
$y{f}^{\prime }(y)=f(y)+y{f}^{\prime }(1)=f(y)+y$.....(1)
Now again differentiate w.r.t. $y$
$y{f}^{\prime \prime }(y)+f^{\prime }(y)=f^{\prime }(y)+1$
$\therefore f^{\prime \prime }(y)=\frac{1}{y}\Rightarrow f^{\prime \prime }\left(\frac{1}{7}\right)=7$