Q101 / 180 · KCET 2021 · Mathematics

If the parametric equation of curve is given by $x=\cos \theta+\log \tan \frac{\theta}{2}$ and $y=\sin \theta$, then the points for which $\frac{d y}{d x}=0$ are given by

  • A
    $\theta=\frac{n \pi}{2}, n \in Z$
  • B
    $\theta=(2 n+1) \frac{\pi}{2}, n \in Z$
  • C
    $\theta=(2 n+1) \pi, n \in Z$
  • Correct answer
    D
    $\theta=n \pi, n \in z$

Solution

$x=\cos \theta+\log \tan \frac{\theta}{2}$ Oñ đifferrēñtiāting ̄ w.r.t. $\theta$, wē gẹt $\frac{d x}{d \theta}=-\sin \theta+\frac{1}{\tan \frac{\theta}{2}} \times \sec ^{2} \frac{\theta}{2} \times \frac{1}{2}$ $\begin{aligned} &=-\sin \theta+\frac{1}{\sin \theta} \\ &=\frac{1-\sin ^{2} \theta}{\sin \theta} \\ &=\frac{\cos ^{2} \theta}{\sin \theta} ...(i) \end{aligned}$ Now, $y=\sin \theta$ On differentiating w.r.t. $\theta$, $\frac{d y}{d \theta}=\cos \theta...(ii)$ $\begin{aligned} \therefore \quad \frac{d y}{d x} &=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\ &=\frac{\cos \theta}{\frac{\cos ^{2} \theta}{\sin \theta}}=\tan \theta \end{aligned}$ [from Eqs. (i) and (ii)] If $\frac{d y}{d x}=0$, then $\tan \theta=0$. Hence, $\theta=n \pi, n \in Z$.

Hence, the correct answer is option: 'D'.

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