यदि f(x) = left| begin{array}{ccc} -sin x & 2 | vedclass

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(D) दिया गया है $f(x) = \left| \begin{array}{ccc} -\sin x & 2 \sin 2x & 4 \cos^2 x \ \cos x & 4 \sin^2 x & 2 \sin 2x \ 0 & -\cos x & \sin x \end{arr

Vedclass · Accepted Answer

$-4$

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(D) दिया गया है $f(x) = \left| \begin{array}{ccc} -\sin x & 2 \sin 2x & 4 \cos^2 x \ \cos x & 4 \sin^2 x & 2 \sin 2x \ 0 & -\cos x & \sin x \end{array} ight|$.प्रथम स्तंभ के अनुदिश सारणिक का विस्तार करने पर:$f(x) = -\sin x (4 \sin^2 x \sin x - (2 \sin 2x)(-\cos x)) - \cos x (2 \sin 2x \sin x - (4 \cos^2 x)(-\cos x)) + 0$$f(x) = -\sin x (4 \sin^3 x + 4 \sin x \cos^2 x) - \cos x (4 \sin^2 x \cos x + 4 \cos^3 x)$$f(x) = -4 \sin^2 x (\sin^2 x + \cos^2 x) - 4 \cos^2 x (\sin^2 x + \cos^2 x)$$f(x) = -4 \sin^2 x - 4 \cos^2 x = -4(\sin^2 x + \cos^2 x) = -4$.चूंकि $f(x) = -4$ एक अचर फलन है,इसका अवकलज $f'(x) = 0$ है।अतः,$f\left(\frac{5\pi}{4}ight) + f'\left(\frac{5\pi}{4}ight) = -4 + 0 = -4$.

यदि $f(x) = \left| \begin{array}{ccc} -\sin x & 2 \sin 2x & 4 \cos^2 x \\ \cos x & 4 \sin^2 x & 2 \sin 2x \\ 0 & -\cos x & \sin x \end{array} \right|$ है,तो $f\left(\frac{5\pi}{4}\right) + f'\left(\frac{5\pi}{4}\right) = $

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