JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}\) and \(y ( t )=2 \sqrt{2} \sin t \sqrt{\sin 2 t }, t \in\left(0, \frac{\pi}{2}\right)\). Then \(\frac{1+\left(\frac{ dy }{ dx }\right)^{2}}{\frac{ d ^{2} y }{ dx ^{2}}}\) at \(t =\frac{\pi}{4}\) is equal to..
- A \(\frac{-2 \sqrt{2}}{3}\)
- B \(\frac{2}{3}\)
- C \(\frac{1}{3}\)
- D \(\frac{-2}{3}\)
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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