JEE Main 2023 — Mathematics Calculus

Given:

$$f(x)=|\begin{matrix}1+{\mathrm{sin}}^{2}x & {\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ {\mathrm{sin}}^{2}x & 1+{\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ {\mathrm{sin}}^{2}x & {\mathrm{cos}}^{2}x & 1+\mathrm{sin}2x\end{matrix}|$$f(x)=∣1+sin2xsin2xsin2x​cos2x1+cos2xcos2x​sin2xsin2x1+sin2x​∣

Applying ${C}_{1}\rightarrow {C}_{1}+{C}_{2}+{C}_{3}$C1​→C1​+C2​+C3​

$$f(x)=|\begin{matrix}2+\mathrm{sin}2x & {\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ 2+\mathrm{sin}2x & 1+{\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ 2+\mathrm{sin}2x & {\mathrm{cos}}^{2}x & 1+\mathrm{sin}2x\end{matrix}|$$f(x)=∣2+sin2x2+sin2x2+sin2x​cos2x1+cos2xcos2x​sin2xsin2x1+sin2x​∣

$$\Rightarrow f(x)=(2+\mathrm{sin}2x)|\begin{matrix}1 & {\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ 1 & 1+{\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ 1 & {\mathrm{cos}}^{2}x & 1+\mathrm{sin}2x\end{matrix}|$$⇒f(x)=(2+sin2x)∣111​cos2x1+cos2xcos2x​sin2xsin2x1+sin2x​∣

Applying ${R}_{2}\rightarrow {R}_{2}-{R}_{1}$R2​→R2​−R1​ and ${R}_{3}\rightarrow {R}_{3}-{R}_{1}$R3​→R3​−R1​

$$f(x)=(2+\mathrm{sin}2x)|\begin{matrix}1 & {\mathrm{cos}}^{2}x & \mathrm{sin}2x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}|$$f(x)=(2+sin2x)∣100​cos2x10​sin2x01​∣

$$\Rightarrow f(x)=(2+\mathrm{sin}2x)(1)=2+\mathrm{sin}2x$$⇒f(x)=(2+sin2x)(1)=2+sin2x

Now, for

$$x\in [\frac{\pi }{6},\frac{\pi }{3}]$$x∈[6π​,3π​]

$$\Rightarrow 2x\in [\frac{\pi }{3},\frac{2\pi }{3}]$$⇒2x∈[3π​,32π​]

$$\Rightarrow \mathrm{sin}2x\in [\frac{\sqrt{3}}{2},1]$$⇒sin2x∈[23​​,1]

$$\Rightarrow 2+\mathrm{sin}2x\in [2+\frac{\sqrt{3}}{2},3]$$⇒2+sin2x∈[2+23​​,3]

Hence,

$$\beta =2+\frac{\sqrt{3}}{2}$$β=2+23​​

$$\alpha =3$$α=3

So,

$${\beta }^{2}-2\alpha =4+\frac{3}{4}+2\sqrt{3}-2\sqrt{3}=\frac{19}{4}$$β2−2α=4+43​+23​−23​=419​