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