Question

The value of $\frac{d^{2}y}{d{x}^2}$ at the point where $f\left(t\right)=g\left(t\right)$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}=a^{t\ln b}$ ---------- (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),
$xy=1$
$f\left(t\right)=g\left(t\right)⟹f\left(t\right)=f(-t)⟹t=0[\because f(t)$ is one-one function$]$
At $t=0,x=y=1$
$\because xy=1$, $\frac{dy}{dx}=\frac{-1}{x^2}$ and $\frac{d^{2}y}{d{x}^2}=\frac{2}{x^3}$
At $x=1$, $\frac{d^{2}y}{d{x}^2}=2$