TS EAMCET · Maths · Application of Derivatives
The equation of the normal drawn to the curve \(y=\sin 3 x\) at \(x=\frac{\pi}{4}\) is
- A \(y=\frac{\sqrt{3}}{2}\left(x+\frac{6-\pi}{4}\right)\)
- B \(y=\frac{\sqrt{2}}{3}\left(x+\frac{6-\pi}{4}\right)\)
- C \(y=\frac{\sqrt{3}}{2}\left(x-\frac{6-\pi}{4}\right)\)
- D \(y=\frac{\sqrt{2}}{3}\left(x-\frac{6-\pi}{4}\right)\)
Answer & Solution
Correct Answer
(B) \(y=\frac{\sqrt{2}}{3}\left(x+\frac{6-\pi}{4}\right)\)
Step-by-step Solution
Detailed explanation
At \(x=\pi / 4, y=1 / \sqrt{2}\) \(\therefore\) Point is \((\pi / 4,1 / \sqrt{2})\) Now, differentiating Eq. (i) w.r.t \(x\), we get…
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