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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
Let $f(\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 \pi+\theta)\right)-2\left(1-\sin ^2 2 \theta\right)$ and $S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$. If $4 \beta=\sum_{\theta \in S} \theta$ then $f(\beta)$ is equal to
  • A
    $\frac{11}{8}$
  • B
    $\frac{5}{4}$
  • C
    $\frac{9}{8}$
  • D
    $\frac{3}{2}$
✓

Answer

$ 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)$
$S =\left\{\theta \in[0, \pi]: f ^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$
$\Rightarrow f (\theta)=3\left(\cos ^4 \theta+\sin ^4 \theta\right)-2 \cos ^2 2 \theta$
$\Rightarrow f (\theta)=3\left(1-\frac{1}{2} \sin ^2 2 \theta\right)-2 \cos ^2 2 \theta$
$\Rightarrow f (\theta)=3-\frac{3}{2} \sin ^2 2 \theta-2 \cos ^2 \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^2 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$f^{\prime}(\theta)=\sin 4 \theta$
$\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}$
$\Rightarrow 4 \theta=n \pi+(-1)^n \frac{\pi}{3}$
$\Rightarrow \theta=\frac{ n \pi}{4}+(-1)^{ n } \frac{\pi}{12}$
$\Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right)$
$\Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2}$
$\Rightarrow \beta=\frac{3 \pi}{8} $
$\Rightarrow f (\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4}$

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