Verified Solution • 2025–26 Session moderate

QJEEM23Ma9362: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9362 Solution: Let and . If , then is equal to

Let f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3π+θ))−2(1−sin⁡22θ)f(\theta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - \theta } \right) + {{\sin }^4}(3\pi + \theta )} \right) - 2(1 - {\sin ^2}2\theta ) and S={θ∈[0,π]:f′(θ)=−32}S = \left\{ {\theta \in [0,\pi ]:f'(\theta ) = - {{\sqrt 3 } \over 2}} \right\} . If 4β=∑θ∈Sθ4\beta = \sum\limits_{\theta \in S} \theta , then f(β)f(\beta ) is equal to

Multiple Choice Options

  • A 98\frac{9}{8}
  • B 32\frac{3}{2}
  • C 54\frac{5}{4}
  • D 118\frac{11}{8}

Step-by-Step Verified Solution

f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3x+θ))−2(1−sin⁡22θ)f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right)

S={θ∈[0,π]:f′(θ)=−32}S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}

⇒f(θ)=3(cos⁡4θ+sin⁡4θ)−2cos⁡22θ\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta

⇒f(θ)=3(1−12sin⁡22θ)−2cos⁡22θ\Rightarrow \mathrm{f}(\theta)=3\left(1-\frac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta

⇒f(θ)=3−32sin⁡22θ−2cos⁡2θ\Rightarrow \mathrm{f}(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta

=32−12cos⁡22θ=32−12(1+cos⁡4θ2)=\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)

f(θ)=54−cos⁡4θ4f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}

f′(θ)=sin⁡4θf^{\prime}(\theta)=\sin 4 \theta

⇒f′(θ)=sin⁡4θ=−32\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}

⇒4θ=nπ+(−1)nπ3\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}

⇒θ=nπ4+(−1)nπ12\Rightarrow \theta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12}

amp;⇒θ=π12,(π4−π12),(π2+π12),(3π4−π12)amp;⇒4β=π4+π2+3π4=3π2amp;⇒β=3π8⇒f(β)=54−cos⁡3π24=54\begin{aligned}& \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\\\& \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\\\& \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4}\end{aligned}

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