Q104 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
If $\sec \left(\log _2 y^2\right)=\operatorname{cosec}\left(\log _2 x^2\right)$, then $\frac{d y}{d x}=$
- A$\frac{x}{y}$
- B$\frac{y}{x}$
- Correct answerC$-\frac{y}{x}$
- D$-\frac{x}{y}$
Solution
$\begin{aligned} & \sec \left(\log _2 y^2\right)=\operatorname{cosec}\left(\log _2 x^2\right) \\ & \Rightarrow \sec \left(2 \log _2 y\right)=\operatorname{cosec}\left(2 \log _2 x\right) \\ & \Rightarrow \sec ^2\left(2 \log _2 y\right)=\operatorname{cosec}^2\left(2 \log _2 x\right) \\ & \Rightarrow \tan ^2\left(2 \log _2 y\right)=\cot ^2\left(2 \log _2 x\right) \\ & \Rightarrow \tan \left(2 \log _2 y\right)=\cot \left(2 \log _2 x\right) \\ & \Rightarrow \frac{\sin \left(2 \log _2 y\right)}{\cos \left(2 \log _2 y\right)}=\frac{\cos \left(2 \log _2 x\right)}{\sin \left(2 \log _2 x\right)} \\ & \Rightarrow \cos \left(2 \log _2 x\right) \cdot \cos \left(2 \log _2 y\right) \\ & \Rightarrow \cos \left(2 \log _2 x+2 \log 2 y\right)=0 \\ & \Rightarrow \cos 2\left(\log _2 x y\right)=0 \Rightarrow 2\left(\log _2 x y\right)=\frac{\pi}{2} \\ & \Rightarrow \frac{\log x y}{\log 2}=\frac{\pi}{4} \Rightarrow \log x y=\frac{\pi}{4} \log _2 \\ & \text { Differentiate both sides } \\ & x \frac{d y}{d x}+y \\ & \frac{x y}{x}=0 \\ & \therefore \quad \frac{d y}{d x}=\frac{-y}{x} .\end{aligned}$
Hence, the correct answer is option: 'C'.