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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

माना \(x ( t )=2 \sqrt{2} \cos t \sqrt{\sin 2 t }\) तथा \(y ( t )=2 \sqrt{2} \sin t \sqrt{\sin 2 t }, t \in\left(0, \frac{\pi}{2}\right)\). हैं।तब \(\frac{1+\left(\frac{ dy }{ dx }\right)^2}{\frac{ d ^2 y }{ dx ^2}}\) पर \(t =\frac{\pi}{4}\) बराबर है।

  1. A \(\frac{-2 \sqrt{2}}{3}\)
  2. B \(\frac{2}{3}\)
  3. C \(\frac{1}{3}\)
  4. D \(\frac{-2}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{-2}{3}\)

Step-by-step Solution

Detailed explanation

\(x =2 \sqrt{2} \cos t \sqrt{\sin 2 t }\) \(\frac{ dx }{ dt }=\frac{2 \sqrt{2} \cos 3 t }{\sqrt{\sin 2 t }}\) \(y ( t )=2 \sqrt{2} \sin t \sqrt{\sin 2 t }\) \(\frac{ dy }{ dt }=\frac{2 \sqrt{2} \sin 3 t }{\sqrt{\sin 2 t }}\) \(\frac{ dy }{ dx }=\tan 3 t\)…
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