JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 10)
Let $$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then
$${\alpha ^2} - {\beta ^2} = 4\sqrt 3 $$
$${\beta ^2} - 2\sqrt \alpha = {{19} \over 4}$$
$${\beta ^2} + 2\sqrt \alpha = {{19} \over 4}$$
$${\alpha ^2} + {\beta ^2} = {9 \over 2}$$
Explanation
$$f(x) = \left| {\matrix{
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$$ \begin{aligned} & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\1 & 1+\cos ^{2} x & \sin 2 x \\1 & \cos ^{2} x & 1+\sin 2 x\end{array}\right| \\\\ & R_{2} \rightarrow R_{2} \rightarrow R_{1} ; R_{3} \rightarrow R_{3} \rightarrow R_{1} \\\\ & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right| \\\\ & f(x)=2+\sin 2 x ; x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] \\\\ & f(x)_{\max }=2+1=3 \text { for } x=\frac{\pi}{4} \\\\ & f(x)_{\min }=2+\frac{\sqrt{3}}{2} \text { for } x=\frac{\pi}{6}, \frac{\pi}{3} \\\\ & \beta^{2}-2 \sqrt{\alpha}=4+\frac{3}{4}+2 \sqrt{3}-2 \sqrt{3} \\\\ & =\frac{19}{4} \end{aligned} $$
$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$$ \begin{aligned} & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\1 & 1+\cos ^{2} x & \sin 2 x \\1 & \cos ^{2} x & 1+\sin 2 x\end{array}\right| \\\\ & R_{2} \rightarrow R_{2} \rightarrow R_{1} ; R_{3} \rightarrow R_{3} \rightarrow R_{1} \\\\ & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right| \\\\ & f(x)=2+\sin 2 x ; x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] \\\\ & f(x)_{\max }=2+1=3 \text { for } x=\frac{\pi}{4} \\\\ & f(x)_{\min }=2+\frac{\sqrt{3}}{2} \text { for } x=\frac{\pi}{6}, \frac{\pi}{3} \\\\ & \beta^{2}-2 \sqrt{\alpha}=4+\frac{3}{4}+2 \sqrt{3}-2 \sqrt{3} \\\\ & =\frac{19}{4} \end{aligned} $$
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