ધારો કે f(theta)=3left(sin ^4left(frac{3 pi}{ | vedclass

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(B) આપેલ છે કે $f(	heta)=3\left(\cos ^4 	heta+\sin ^4 	hetaight)-2 \cos ^2 2 	heta$. $\sin ^4 	heta+\cos ^4 	heta = 1-2 \sin ^2 	heta \cos ^2

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$\frac{5}{4}$

Vedclass · Answer

(B) આપેલ છે કે $f(	heta)=3\left(\cos ^4 	heta+\sin ^4 	hetaight)-2 \cos ^2 2 	heta$. $\sin ^4 	heta+\cos ^4 	heta = 1-2 \sin ^2 	heta \cos ^2 	heta = 1-\frac{1}{2} \sin ^2 2 	heta$ નો ઉપયોગ કરતા,આપણને મળે: $f(	heta)=3\left(1-\frac{1}{2} \sin ^2 2 	hetaight)-2 \cos ^2 2 	heta = 3-\frac{3}{2} \sin ^2 2 	heta-2 \cos ^2 2 	heta$. $\sin ^2 2 	heta = 1-\cos ^2 2 	heta$ હોવાથી: $f(	heta)=3-\frac{3}{2}(1-\cos ^2 2 	heta)-2 \cos ^2 2 	heta = \frac{3}{2}-\frac{1}{2} \cos ^2 2 	heta$. $\cos ^2 2 	heta = \frac{1+\cos 4 	heta}{2}$ નો ઉપયોગ કરતા: $f(	heta)=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 	heta}{2}ight) = \frac{5}{4}-\frac{\cos 4 	heta}{4}$. હવે,$f^{\prime}(	heta) = \frac{d}{d 	heta} \left(\frac{5}{4}-\frac{\cos 4 	heta}{4}ight) = \sin 4 	heta$. આપેલ છે કે $f^{\prime}(	heta) = -\frac{\sqrt{3}}{2}$,તેથી $\sin 4 	heta = -\frac{\sqrt{3}}{2}$. $	heta \in [0, \pi]$ માટે,$4 	heta \in [0, 4 \pi]$. $\sin 4 	heta = -\frac{\sqrt{3}}{2}$ ના ઉકેલો $4 	heta = \frac{4 \pi}{3}, \frac{5 \pi}{3}, \frac{10 \pi}{3}, \frac{11 \pi}{3}$ છે. તેથી,$	heta \in \left\{\frac{\pi}{3}, \frac{5 \pi}{12}, \frac{5 \pi}{6}, \frac{11 \pi}{12}ight\}$. $S$ માં રહેલા $	heta$ નો સરવાળો $4 \beta = \frac{\pi}{3}+\frac{5 \pi}{12}+\frac{5 \pi}{6}+\frac{11 \pi}{12} = \frac{4 \pi+5 \pi+10 \pi+11 \pi}{12} = \frac{30 \pi}{12} = \frac{5 \pi}{2}$. તેથી $\beta = \frac{5 \pi}{8}$. $f(\beta) = \frac{5}{4}-\frac{\cos(4 \cdot \frac{5 \pi}{8})}{4} = \frac{5}{4}-\frac{\cos(5 \pi / 2)}{4} = \frac{5}{4}-0 = \frac{5}{4}$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

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