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QJEEM23Ma544f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma544f Solution: Let f ( x ) =

Let f(x)=|1+sin2xcos2xsin⁡2xsin2x1+cos2xsin⁡2xsin2xcos2x1+sin⁡2x|,x∈[π6,π3]. If α\alpha and β\beta respectively are the maximum and the minimum values of ff , then

Multiple Choice Options

  • A α2−β2=43{\alpha ^2} - {\beta ^2} = 4\sqrt 3
  • B β2−2α=194{\beta ^2} - 2\sqrt \alpha = {{19} \over 4}
  • C β2+2α=194{\beta ^2} + 2\sqrt \alpha = {{19} \over 4}
  • D α2+β2=92{\alpha ^2} + {\beta ^2} = {9 \over 2}

Step-by-Step Verified Solution

f(x)=|1+sin2xcos2xsin⁡2xsin2x1+cos2xsin⁡2xsin2xcos2x1+sin⁡2x|

C1→C1+C2+C3C_{1} \rightarrow C_{1}+C_{2}+C_{3}

amp;=(2+sin⁡2x)∣1amp;cos⁡2xamp;sin⁡2x1amp;1+cos⁡2xamp;sin⁡2x1amp;cos⁡2xamp;1+sin⁡2x∣amp;R2→R2→R1;R3→R3→R1amp;=(2+sin⁡2x)∣1amp;cos⁡2xamp;sin⁡2x0amp;1amp;00amp;0amp;1∣amp;f(x)=2+sin⁡2x;x∈[π6,π3]amp;f(x)max⁡=2+1=3 for x=π4amp;f(x)min⁡=2+32 for x=π6,π3amp;β2−2α=4+34+23−23amp;=194\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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