Question

A curve is represented parametrically by the equations $x=f\left(t\right)=a^{\ln \left(b^t\right)}$ and $y=g\left(t\right)=b^{-\ln \left(a^t\right)};a,b>0$ and $a\ne 1,b\ne 1$ where $t\in R$.

The value of $\frac{f\left(t\right)}{f^{\prime }\left(t\right)}\cdot \frac{f^{\prime \prime }(-t)}{f^{\prime }(-t)}+\frac{f(-t)}{f^{\prime }(-t)}\cdot \frac{f^{\prime \prime }\left(t\right)}{f^{\prime }\left(t\right)}\forall t\in R$, equal to

• A. $-2$
• B. $2$
• C. $-4$
• D. $4$

Answer 1

The correct option is B $2$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}\Rightarrow x=a^{t\ln b}={\left(a^{\ln b}\right)}^t$
$y=g\left(t\right)=b^{-\ln \left(a^t\right)}\Rightarrow y=b^{-t\ln a}={\left(b^{\ln a}\right)}^{-t}\Rightarrow y={\left(a^{\ln b}\right)}^{-t}\Rightarrow y=a^{-t\ln b}=\frac{1}{x}\Rightarrow xy=1$
$\Rightarrow fg=1\Rightarrow f{g}^{\prime }+f^{\prime }g=0\Rightarrow f{g}^{\prime \prime }+f^{\prime }{g}^{\prime }+f^{\prime }{g}^{\prime }+f^{\prime \prime }g=0\Rightarrow f{g}^{\prime \prime }+f^{\prime \prime }g=-2f^{\prime }{g}^{\prime }\cdots \left(1\right)$

We know that,
$f(-t)=g\left(t\right)\Rightarrow g^{\prime }\left(t\right)=-f^{\prime }(-t)\Rightarrow g^{\prime \prime }\left(t\right)=f^{\prime \prime }(-t)$

Putting in equation $\left(1\right)$,
$f\left(t\right)f^{\prime \prime }(-t)+f^{\prime \prime }\left(t\right)f(-t)=2f^{\prime }\left(t\right)f^{\prime }(-t)\Rightarrow \frac{f\left(t\right)}{f^{\prime }\left(t\right)}\cdot \frac{f^{\prime \prime }(-t)}{f^{\prime }(-t)}+\frac{f(-t)}{f^{\prime }(-t)}\cdot \frac{f^{\prime \prime }\left(t\right)}{f^{\prime }\left(t\right)}=2$