Question

The function $f:R\to R$ satisfies $f\left(x^2\right).f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right).f^{\prime }\left(x^2\right)$ for all real $x.$ Given that $f\left(1\right)=1$ and $f^{\prime \prime \prime }\left(1\right)=8,$ compute the value of $f^{\prime }\left(1\right)+f^{\prime \prime }\left(1\right).$

• A. $6$
• B. $2$
• C. $7$
• D. None of these

Answer 1

Answer: (A)

$6$

$f\left(x^2\right).f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right).f^{\prime }\left(x^2\right)$ ...(1)

$\Rightarrow 2x{f}^{\prime }\left(x^2\right)f^{\prime \prime }\left(x\right)+f\left(x^2\right)f^{\prime \prime \prime }\left(x\right)=f^{\prime \prime }\left(x\right)f^{\prime }\left(x^2\right)+2f^{\prime \prime }\left(x^2\right)f^{\prime }\left(x\right)$ ...(2)

Putting $x=1$ in equation (2), we get

$2f^{\prime }\left(1\right)f^{\prime \prime }\left(1\right)+f\left(1\right)f^{\prime \prime \prime }\left(1\right)=f^{\prime \prime }\left(1\right)+f\left(1\right)+f^{\prime \prime }\left(1\right)$

$\Rightarrow 2f^{\prime }\left(1\right)f^{\prime \prime }\left(1\right)+1\times 8=3f^{\prime \prime }\left(1\right)f^{\prime }\left(1\right)$

$\Rightarrow f^{\prime }\left(1\right)f^{\prime \prime }\left(1\right)=8$ ...(3)

Putting $x=1$ in (1), we get,

$f\left(1\right)f^{\prime \prime }\left(1\right)=f^{\prime }\left(1\right)f^{\prime }\left(1\right)$

$\Rightarrow f^{\prime \prime }\left(1\right)={\left(f^{\prime }\left(1\right)\right)}^2$

$\Rightarrow f^{\prime }\left(1\right).{\left(f^{\prime }\left(1\right)\right)}^2=8\Rightarrow f^{\prime }\left(1\right)=2\Rightarrow f^{\prime \prime }\left(1\right)=4\Rightarrow f^{\prime }\left(1\right)+f^{\prime \prime }\left(1\right)=6$