QJEEM21Maff01: JEE Main 2021 Mathematics - Class 12 JEEM21Maff01 Solution: Let f ( x ) =

Question

Verified Solution • 2025–26 Session moderate

Let

f (x) = | \begin{matrix} \sin^{2} x & - 2 + \cos^{2} x & \cos 2 x \\ 2 + \sin^{2} x & \cos^{2} x & \cos 2 x \\ \sin^{2} x & \cos^{2} x & 1 + \cos 2 x \end{matrix} |, x \in [0, π]

. Then the maximum value of f(x) is equal to ______________.

Accepted Answer

| \begin{matrix} - 2 & - 2 & 0 \\ 2 & 0 & - 1 \\ \sin^{2} x & \cos^{2} x & 1 + \cos 2 x \end{matrix} | (\begin{matrix} R_{1} \to R_{1} - R_{2} \\ & R_{2} \to R_{2} - R_{3} \end{matrix})

=

- 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)

=

4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)

\therefore

f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}

\Rightarrow

f{(x)_{\max }} = 4 + 2 = 6

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