Q92 / 155 · WBJEE 2023 · Mathematics
If \(x=\sin \theta\) and \(y=\sin k \theta\), then \(\left(1-x^2\right) y_2-x y_1-\alpha y=0\), for \(\alpha=\)
- A\(\mathrm{k}\)
- B\(-\mathrm{k}\)
- Correct answerC\(-\mathrm{k}^2\)
- D\(\mathrm{k}^2\)
Solution
Hint : \(x=\sin \theta\)
\(y=\sin k \theta\)
\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{k \cos k \theta}{\cos \theta}=y_1 \ldots\ldots (I)\)
\(\begin{aligned}
& y_2=\frac{d^2 y}{d x^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}=\left[\frac{-k^2 \sin (k \theta) \cdot \cos \theta+k \cos (k \theta) \sin \theta}{\cos ^2 \theta}\right] \times \frac{1}{\cos \theta} \\
& y_2=\frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cos \theta}{\cos ^3 \theta} \ldots \ldots \text { (II) }
\end{aligned}\)
also, \(1-x^2=1-\sin ^2 \theta=\cos ^2 \theta \ldots \ldots \text { (III) }\)
So, putting the values of \(1-x^2, y_1\) and \(y_2\) from (I), (II), (III) we get
\(\begin{aligned}
& \left(1-x^2\right) y_2-x y_1=\alpha y \\
& \Rightarrow\left(\cos ^2 \theta\right)\left[\frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cdot \cos \theta}{\cos ^3 \theta}\right]-\frac{k \sin \theta \cos (k \theta)}{\cos \theta}=\alpha(\sin k \theta) \\
& \Rightarrow \frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cos \theta-k \sin \theta \cos (k \theta)}{\cos \theta}=\alpha \sin (k \theta) \\
& \Rightarrow-k^2 \sin (k \theta)=\alpha \sin (k \theta)
\end{aligned}\)
So, \(\alpha=-k^2\)
Hence, the correct answer is option: 'C'.