Verified Solution • 2025–26 Session moderate

QJEEM23Mafcd0: JEE Main 2023 Mathematics - Class 11 JEEM23Mafcd0 Solution: Let and β = ∑ x ∈ S tan 2 ⁡ ( x 3 ) , then is equal to :

Let S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\} and β=∑x∈Stan2⁡(x3), then 16(β−14)2\frac{1}{6}(\beta-14)^{2} is equal to :

Multiple Choice Options

  • A 16
  • B 32
  • C 8
  • D 64

Step-by-Step Verified Solution

We have, S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\} and

β=∑x∈Stan2⁡(x3)

amp;91−tan⁡2x+9tan⁡2x=10amp;⇒99tan⁡2x+9tan⁡2x=10\begin{aligned}& 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10 \\\\&\Rightarrow \frac{9}{9^{\tan ^2 x}}+9^{\tan ^2 x}=10\end{aligned}

Let 9tan⁡2x=t9^{\tan ^2 x}=t

9t+t=10⇒t2−10t+9=0⇒t=9 or 1\frac{9}{t}+t=10 \Rightarrow t^2-10 t+9=0 \Rightarrow t=9 \text { or } 1

If t=9,9tan⁡2x=9⇒x=±π4t=9,9^{\tan ^2 x}=9 \Rightarrow x= \pm \frac{\pi}{4}

If t=1,9tan⁡2x=1⇒x=0t=1,9^{\tan ^2 x}=1 \Rightarrow x=0

Also,

βamp;=∑x∈Stan⁡2(x3)amp;=tan⁡2(03)+tan⁡2(π12)+tan⁡2(−π12)amp;=0+2tan⁡2π12=2[2−3]2amp;=2[4+3−43]=14−83\begin{aligned}\beta & =\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right) \\\\& =\tan ^2\left(\frac{0}{3}\right)+\tan ^2\left(\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right) \\\\& =0+2 \tan ^2 \frac{\pi}{12}=2[2-\sqrt{3}]^2 \\\\& =2[4+3-4 \sqrt{3}]=14-8 \sqrt{3}\end{aligned}

 So, now 16(β−14)2=16[14−83−14]2=16×64×3=32\text { So, now } \frac{1}{6}(\beta-14)^2=\frac{1}{6}[14-8 \sqrt{3}-14]^2=\frac{1}{6} \times 64 \times 3=32

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