CUET · MATHS · PYQ PAPER 2025
Let \(y=\sin \left(\cos x^2\right)\), then the value of \(\frac{d y}{d x}\) at \(x=\frac{\sqrt{\pi}}{2}\) is equal to:
- A \(-\frac{\sqrt{\pi}}{2} \cos \left(\frac{1}{\sqrt{2}}\right)\)
- B \(-\sqrt{\frac{\pi}{2}} \cos \left(\frac{1}{\sqrt{2}}\right)\)
- C \(-\sqrt{\frac{\pi}{2}} \sin \left(\frac{1}{\sqrt{2}}\right)\)
- D \(\sqrt{\frac{\pi}{2}} \sin \left(\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(B) \(-\sqrt{\frac{\pi}{2}} \cos \left(\frac{1}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x} = -2x \sin x^2 \cos \left(\cos x^2\right)\) \(\frac{d y}{d x}\Big|_{x=\frac{\sqrt{\pi}}{2}} = -2\left(\frac{\sqrt{\pi}}{2}\right) \sin\left(\left(\frac{\sqrt{\pi}}{2}\right)^2\right) \cos\left(\cos\left(\left(\frac{\sqrt{\pi}}{2}\right)^2\right)\right)\)…
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