JEE 2023 — Mathematics PYQ

Solution:

Simplify $f(\theta )$:

$\sin (\frac{3\pi }{2}-\theta )=-\cos \theta ⟹{\sin }^4={\cos }^4\theta$.

$\sin (3\pi +\theta )=-\sin \theta ⟹{\sin }^4={\sin }^4\theta$.

$f(\theta )=3({\cos }^4\theta +{\sin }^4\theta )-2{\cos }^{2}2\theta$.

Using ${\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{\sin }^{2}2\theta$:

$f(\theta )=3(1-\frac{1}{2}{\sin }^{2}2\theta )-2{\cos }^{2}2\theta =3-\frac{3}{2}{\sin }^{2}2\theta -2(1-{\sin }^{2}2\theta )=1+\frac{1}{2}{\sin }^{2}2\theta$.

Differentiating: $f^{\prime }(\theta )=\frac{1}{2}(2\sin 2\theta \cdot \cos 2\theta \cdot 2)=\sin 4\theta$.

Given $\sin 4\theta =-\frac{\sqrt{3}}{2}$ for $\theta \in [0,\pi ]⟹4\theta \in [0,4\pi ]$.

Possible values for $4\theta$: $\frac{4\pi }{3},\frac{5\pi }{3},\frac{10\pi }{3},\frac{11\pi }{3}$.

Sum of $\theta =\frac{1}{4}\left(\frac{4\pi +5\pi +10\pi +11\pi }{3}\right)=\frac{30\pi }{12}=\frac{5\pi }{2}$.

Since $4\beta =\sum \theta ⟹4\beta =\frac{5\pi }{2}⟹2\beta =\frac{5\pi }{4}$.

$f(\beta )=1+\frac{1}{2}{\sin }^2(2\beta )=1+\frac{1}{2}{\sin }^2(\frac{5\pi }{4})=1+\frac{1}{2}(\frac{1}{2})=\frac{5}{4}$.

Correct Option: (1)