JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the tangent to the curve \(y\, = \,\frac{x}{{{x^2}\, - \,3}},\,x\, \in \,R,\,(x\, \ne \, \pm \,\sqrt 3 )\) at a point \((\alpha ,\,\beta )\,\ne\,(0,0)\) on it is parallel to the line \(2x + 6y -11 = 0\) then
- A \(\left| {2\alpha \, + \,6\beta } \right|\, = \,11\)
- B \(\left| {2\alpha \, + \,6\beta } \right|\, = \,19\)
- C \(\left| {6\alpha \, + \,2\beta } \right|\, = \,19\)
- D \(\left| {6\alpha \, + \,2\beta } \right|\, = \,9\)
Answer & Solution
Correct Answer
(C) \(\left| {6\alpha \, + \,2\beta } \right|\, = \,19\)
Step-by-step Solution
Detailed explanation
\({\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\alpha ,\beta } \right)}} = \frac{{ - {\alpha ^2} - 3}}{{{{\left( {{\alpha ^2} - 3} \right)}^2}}}\) Given that: \(\frac{{ - {\alpha ^2} - 3}}{{{{\left( {{\alpha ^2} - 3} \right)}^2}}} = - \frac{1}{3}\)…
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