MHT CET · Maths · Application of Derivatives
If the tangent to the curve \(y=\frac{x}{x^2-3}, x \in R,(x \neq \pm \sqrt{3})\) at a point \((\alpha, \beta) \neq(0,0)\) on it, is parallel to the line \(2 x+6 y-11=0\), then
- A \(|2 \alpha+6 \beta|=11\)
- B \(|6 \alpha+2 \beta|=9\)
- C \(|6 \alpha+2 \beta|=19\)
- D \(|2 \alpha+6 \beta|=19\)
Answer & Solution
Correct Answer
(C) \(|6 \alpha+2 \beta|=19\)
Step-by-step Solution
Detailed explanation
\(y=\frac{x}{x^2-3} \Rightarrow \frac{d y}{d x}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2}\)
Now slope of \(2 x+6 y-11=0\) is \(\frac{-1}{3}\)
\(\begin{aligned} & \frac{\mathrm{A}}{\mathrm{Q}}-\frac{1}{3}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2} \\ & \Rightarrow x^4-9 x^2=0 \\ & \Rightarrow x^2\left(x^2-9\right)=0 \\ & \Rightarrow x=0 \text { or } x= \pm 3\end{aligned}\)
But \(x \neq 0\) so \(x= \pm 3 \Rightarrow y= \pm \frac{1}{2}\)
Hence, \(\alpha= \pm 3, \beta= \pm \frac{1}{2}\)
\(\Rightarrow|6 \alpha+2 \beta|=\left| \pm\left(6 \times 3+2 \times \frac{1}{2}\right)\right|=19\)
Now slope of \(2 x+6 y-11=0\) is \(\frac{-1}{3}\)
\(\begin{aligned} & \frac{\mathrm{A}}{\mathrm{Q}}-\frac{1}{3}=\frac{-\left(3+x^2\right)}{\left(x^2-3\right)^2} \\ & \Rightarrow x^4-9 x^2=0 \\ & \Rightarrow x^2\left(x^2-9\right)=0 \\ & \Rightarrow x=0 \text { or } x= \pm 3\end{aligned}\)
But \(x \neq 0\) so \(x= \pm 3 \Rightarrow y= \pm \frac{1}{2}\)
Hence, \(\alpha= \pm 3, \beta= \pm \frac{1}{2}\)
\(\Rightarrow|6 \alpha+2 \beta|=\left| \pm\left(6 \times 3+2 \times \frac{1}{2}\right)\right|=19\)
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