JEE Main 2023 (Online) 29th January Morning ShiftQuestion 1
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 $S=\left\{\theta\in[0,\pi]:f'(\theta)=-\frac{\sqrt3}{2}\right\}$. If $4\beta=\sum_{\theta\in S}\theta$ then $f(\beta)$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Reduce the angles; $f$ simplifies to $1+\frac12\sin^22\theta$, so $f'=\sin4\theta$. Formulas used. (1) $\sin\left(\frac{3\pi}2-\theta\right)=-\cos\theta$, $\sin(3\pi+\theta)=-\sin\theta$(2) $\sin^4\theta+\cos^4\theta=1-\frac12\sin^22\theta$Step 1 - Simplify. By (1) and (2), $$f=3\left(1-\tfrac12\sin^22\theta\right)-2+2\sin^22\theta=1+\tfrac12\sin^22\theta.$$ Step 2 - Derivative. $f'(\theta)=\frac12\cdot2\sin2\theta\cos2\theta\cdot2=\sin4\theta$. Step 3 - Solve $\sin4\theta=-\frac{\sqrt3}{2}$. With $4\theta\in[0,4\pi]$: $$4\theta=\frac{4\pi}3,\ \frac{5\pi}3,\ \frac{10\pi}3,\ \frac{11\pi}3.$$ The $\theta$ values add up to $\frac{30\pi}{12}=\frac{5\pi}{2}$, so $\beta=\frac{5\pi}{8}$. Step 4 - Value. $f(\beta)=1+\frac12\sin^2\frac{5\pi}4=1+\frac14=\frac54$. Answer: (B)