JEE Mains · Maths · STD 12 - 6. Application of derivatives
The equation of a normal to the curve \(\sin \,y = x\,\sin \,\left( {\frac{\pi }{3} + y} \right)\) at \(x\, = 0\), is
- A \(2x - \sqrt 3 y = 0\)
- B \(2x + \sqrt 3 y = 0\)
- C \(2y - \sqrt 3 x = 0\)
- D \(2y + \sqrt 3 x = 0\)
Answer & Solution
Correct Answer
(B) \(2x + \sqrt 3 y = 0\)
Step-by-step Solution
Detailed explanation
Given curve is \(\sin y = x\sin \left( {\frac{\pi }{3} + y} \right)\) Diffwith respect to \(x\), we get \(\cos y\frac{{dy}}{{dx}} = \sin \left( {\frac{\pi }{3} + y} \right) + x\cos \left( {\frac{\pi }{3} + y} \right)\frac{{dy}}{{dx}}\)…
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