Question

The functions $f,g$ and $h$ satisfy the relations $f^{{}^{\prime }}\left(x\right)=g(x+1)$ and $g^{{}^{\prime }}\left(x\right)=h(x-1)$. Then, $f^{{}^{\prime \prime }}\left(2x\right)$ is equal to

• A. $h\left(2x\right)$
• B. $4h\left(2x\right)$
• C. $h(2x-1)$
• D. $h(2x+1)$
• E. $2h^{{}^{\prime }}\left(2x\right)$

Answer 1

Answer: (C)

$h(2x-1)$

We have $f^{{}^{\prime }}\left(x\right)=g(x+1)$
$\Rightarrow f^{{}^{\prime \prime }}\left(x\right)=g^{{}^{\prime }}(x+1)$
But $g^{{}^{\prime }}\left(x\right)=h(x-1)$
$\Rightarrow g^{{}^{\prime }}(x+1)=h(x+1-1)$
$=h\left(x\right)$
Therefore, $f^{{}^{\prime \prime }}\left(x\right)=h\left(x\right)$
$\Rightarrow f^{{}^{\prime \prime }}\left(2x\right)=h\left(2x\right)$