Question

The value of $\frac{f\left(t\right)}{f^{\prime }\left(t\right)}.\frac{f^{\prime \prime }(-t)}{f^{\prime }(-t)}-\frac{f(-t)}{f^{\prime }(-t)}.\frac{f^{\prime \prime }\left(t\right)}{f^{\prime }\left(t\right)}\forall tϵR$ is equal to

• A. $-2$
• B. $2$
• C. $-4$
• D. $4$

Answer 1

The correct option is B $2$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}=a^{t\ln b}$ --------(1)
$y=g\left(t\right)=b^{-\ln \left(a^t\right)}={\left(b^{\ln a}\right)}^{-t}={\left(a^{\ln b}\right)}^{-t}=a^{-t\ln b}$
$\therefore y=g\left(t\right)=a^{\ln \left(b^{-t}\right)}=f(-t)$ --------(2)
From equations (1) and (2),
$xy=1$
$\because xy=1$
$fg=1$
$\therefore f{g}^{\prime }+g{f}^{\prime }=0$
$\therefore f{g}^{\prime \prime }+g^{\prime }{f}^{\prime }+g^{\prime }{f}^{\prime }+g{f}^{\prime \prime }=0$
or $f{g}^{\prime \prime }+g{f}^{\prime \prime }+2g^{\prime }{f}^{\prime }=0$
or $\frac{f}{f^{\prime }}\frac{g^{\prime \prime }}{g^{\prime }}+\frac{g{f}^{\prime \prime }}{g^{\prime }{f}^{\prime }}=-2$ -------(3)
From equation (2), $g\left(t\right)=f(-t)$
$\therefore g^{\prime }\left(t\right)=-f^{\prime }(-t)$
and $g^{\prime \prime }\left(t\right)=f^{\prime \prime }(-t)$
Substituting in equation (3), we get
$\frac{f\left(t\right)}{f^{\prime }\left(t\right)}.\frac{f^{\prime \prime }(-t)}{-f^{\prime }(-t)}-\frac{f(-t)}{f^{\prime }(-t)}.\frac{f^{\prime \prime }\left(t\right)}{f^{\prime }\left(t\right)}=-2$
$\frac{f\left(t\right)}{f^{\prime }\left(t\right)}.\frac{f^{\prime \prime }(-t)}{f^{\prime }(-t)}-\frac{f(-t)}{-f^{\prime }(-t)}.\frac{f^{\prime \prime }\left(t\right)}{f^{\prime }\left(t\right)}=2$