Determinants Question 7
Question 7 - 2024 (30 Jan Shift 1)
If $f(x)=\left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3+2 \cos ^{4} x & 2 \sin ^{4} x & \sin ^{2} 2 x \\ 2 \cos ^{4} x & 3+2 \sin ^{4} x & \sin ^{2} 2 x\end{array}\right|$ then
$\frac{1}{5} f^{\prime}(0)$ is equal to
(1) 0
(2) 1
(3) 2
(4) 6
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Answer (1)
Solution
$\left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3+2 \cos ^{4} x & 2 \sin ^{4} x & \sin ^{2} 2 x \\ 2 \cos ^{4} x & 3+2 \sin ^{2} 4 x & \sin ^{2} 2 x\end{array}\right|$
$R _2 \rightarrow R _2-R _1, R _3 \rightarrow R _3-R _1$
$\left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3\end{array}\right|$
$f(x)=45$
$f^{\prime}(x)=0$
