Question

A function is represented parametrically by the equations $x=\log |2t|,y=|{\tan }^{-1}|t||,t<0.$ Then the value of $\left|\right(1+t^2{)}^2[\frac{1}{t}\frac{d^{2}y}{d{x}^2}-{\left(\frac{dy}{dx}\right)}^2]|$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is B $1$

$x=\log |2t|$
$x=\log (-2t)(\therefore t<0)$
$y=|{\tan }^{-1}|t||={\tan }^{-1}(-t)=-{\tan }^{-1}\left(t\right)$
$\frac{dx}{dt}=\frac{1}{-2t}\times (-2)=\frac{1}{t}\frac{dy}{dt}=-\frac{1}{1+t^2}$

$\frac{dy}{dx}=\frac{\frac{-1}{1+t^2}}{\frac{1}{t}}=\frac{-t}{1+t^2}$
$\frac{d^{2}y}{d{x}^2}=\frac{(1+t^2)\times (-1)\frac{dt}{dx}-(-t)\times \left(2t\right)\frac{dt}{dx}}{(1+t^2{)}^2}=\frac{t}{(1+t^2{)}^2}\times (t^2-1)$

$\left|\right(1+t^2{)}^2[\frac{1}{t}\frac{d^{2}y}{d{x}^2}-{\left(\frac{dy}{dx}\right)}^2]|=\left|\right(1+t^2{)}^2[\frac{t^2-1}{(1+t^2{)}^2}-{\left(\frac{t}{1+t^2}\right)}^2]|=\left|\right(1+t^2{)}^2\times \frac{-1}{(1+t^2{)}^2}|=1$