JEE Main 2024 — 5 April, Shift 2 — Question 19

$\mathrm{y}=\frac{2 \cos \theta+2 \cos ^{2} \theta-1}{4 \cos ^{3} \theta-3 \cos \theta+8 \cos ^{2} \theta-4+5 \cos \theta+2}$

$y=\frac{\left(2 \cos ^{2} \theta+2 \cos \theta-1\right)}{\left(2 \cos ^{2} \theta+2 \cos \theta-1\right)(2 \cos \theta+2)}$

$\mathrm{y}=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right)$

$\Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}$

$y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^{2}} \times(-\sin \theta)\right)$

$\Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}$

$$\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^{2}-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^{4}}\right] &\end{aligned}$$

$\Rightarrow \theta=\frac{\pi}{2} \quad y=1$