જો f(x) = begin{vmatrix} 2 cos^4 x & 2 sin^4 | vedclass

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(A) આપેલ નિશ્ચાયક $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \ 2 \cos^4 x & 3 + 2 \si

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(A) આપેલ નિશ્ચાયક $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$ છે.હરોળ પ્રક્રિયાઓ $R_2 ightarrow R_2 - R_1$ અને $R_3 ightarrow R_3 - R_1$ લાગુ પાડતા:$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \ 3 & 0 & -3 \ 0 & 3 & -3 \end{vmatrix}$.બીજી હરોળને અનુલક્ષીને વિસ્તરણ કરતા:$f(x) = -3 \begin{vmatrix} 2 \sin^4 x & 3 + \sin^2 2x \ 3 & -3 \end{vmatrix} + 0 - (-3) \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x \ 0 & 3 \end{vmatrix}$.$f(x) = -3(-6 \sin^4 x - 3(3 + \sin^2 2x)) + 3(6 \cos^4 x)$.$f(x) = 18 \sin^4 x + 27 + 9 \sin^2 2x + 18 \cos^4 x$.કારણ કે $\sin^2 2x = 4 \sin^2 x \cos^2 x$,તેથી $f(x) = 18(\sin^4 x + \cos^4 x) + 9(4 \sin^2 x \cos^2 x) + 27$.$\sin^4 x + \cos^4 x = 1 - 2 \sin^2 x \cos^2 x$ નો ઉપયોગ કરતા:$f(x) = 18(1 - 2 \sin^2 x \cos^2 x) + 36 \sin^2 x \cos^2 x + 27$.$f(x) = 18 - 36 \sin^2 x \cos^2 x + 36 \sin^2 x \cos^2 x + 27 = 45$.$f(x) = 45$ એ અચળ હોવાથી,$f'(x) = 0$ થાય.તેથી,$\frac{1}{5} f'(0) = 0$.

જો $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$ હોય,તો $\frac{1}{5} f'(0)$ ની કિંમત શોધો.

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