Q54 / 150 · MHT CET 2007 · Mathematics
The slope of the tangent at $(x, y)$ to a curve passing through $\left(1, \frac{\pi}{4}\right)$ is given by $\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)$, then the equation of the curve is
- A$y=\tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]$
- B$y=x \tan ^{-1}\left[\log \left(\frac{x}{e}\right)\right]$
- Correct answerC$y=x \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]$
- DNone of the above
Solution
According to the given condition,
$$
\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)
$$
On putting $y=v x$
$$
\Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}, \text { we get }
$$
$$
\begin{aligned}
& v+x \frac{d v}{d x}=v-\cos ^{2} v \\
\Rightarrow & \frac{d v}{\cos ^{2} v}=-\frac{d x}{x} \\
\Rightarrow & \sec ^{2} v d v=\frac{-1}{x} d x
\end{aligned}
$$
On integrating both sides, we get
$$
\tan v=-\log x+\log c
$$
$$
\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log c
$$
Since, this curve is passing through $(1, \pi / 4)$.
$$
\begin{array}{l}
\therefore \quad \tan \left(\frac{\pi}{4}\right)=-\log 1+\log c \Rightarrow \log c=1 \\
\therefore \quad \tan \left(\frac{y}{x}\right)=-\log x+1 \\
\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log e \\
\Rightarrow \quad \quad y=x \tan ^{-1}\left[\log \left(\frac{e}{x}\right)\right]
\end{array}
$$
Hence, the correct answer is option: 'C'.