Q103 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

If $y=\frac{\alpha x+\beta}{\gamma x+\delta}$, then $2 y_1 y_3=$

  • A
    $2 y_2^3$
  • Correct answer
    B
    $3 y_2^2$
  • C
    $y_2^2$
  • D
    $3 y_3^2$

Solution

$\begin{aligned} & \text { } y=\frac{\alpha x+\beta}{\gamma x+\delta} \\ & \Rightarrow y_1=\frac{\alpha(\gamma x+\delta)-\gamma(\alpha x+\beta)}{(\gamma x+\delta)^2}=\frac{\alpha \delta-\gamma \beta}{(\gamma x+\delta)^2}\end{aligned}$ $\begin{aligned} & \Rightarrow y_2=\frac{-2(\alpha \delta-\gamma \beta) \gamma}{(\gamma x+\delta)^3} \Rightarrow y_3=\frac{6 \gamma^2(\alpha \delta-\gamma \beta)}{(\gamma x+\delta)^4} \\ & \Rightarrow 2 y_1 y_3=\frac{12 \gamma^2(\alpha \delta-\gamma \beta)^2}{(\gamma x+\delta)^6}=3 y_2^2\end{aligned}$

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

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