If $$y(\theta) = \frac{2\cos\theta + \cos 2\theta}{\cos 3\theta + 4\cos 2\theta + 5\cos\theta + 2}$$, then at $$\theta = \frac{\pi}{2}$$, $$y'' + y' + y$$ is equal to :

Write the function as a quotient:
$$y(\theta)=\frac{N(\theta)}{D(\theta)}$$ where

$$N(\theta)=2\cos\theta+\cos2\theta$$
$$D(\theta)=\cos3\theta+4\cos2\theta+5\cos\theta+2$$

First derivatives:

$$N'(\theta)=\frac{d}{d\theta}\!\left(2\cos\theta+\cos2\theta\right)=-2\sin\theta-2\sin2\theta$$
$$D'(\theta)=\frac{d}{d\theta}\!\left(\cos3\theta+4\cos2\theta+5\cos\theta+2\right)=-3\sin3\theta-8\sin2\theta-5\sin\theta$$

Second derivatives:

$$N''(\theta)=\frac{d}{d\theta}\!\left(-2\sin\theta-2\sin2\theta\right)=-2\cos\theta-4\cos2\theta$$
$$D''(\theta)=\frac{d}{d\theta}\!\left(-3\sin3\theta-8\sin2\theta-5\sin\theta\right)=-9\cos3\theta-16\cos2\theta-5\cos\theta$$

At $$\theta=\frac{\pi}{2}$$ the trigonometric values are:
$$\cos\frac{\pi}{2}=0,\;\sin\frac{\pi}{2}=1$$
$$\cos\pi=-1,\;\sin\pi=0$$
$$\cos\frac{3\pi}{2}=0,\;\sin\frac{3\pi}{2}=-1$$

Hence

$$N=-1,\;D=-2$$
$$N'=-2,\;D'=-2$$
$$N''=4,\;D''=16$$

Using the quotient rule $$y'=\dfrac{N'D-ND'}{D^{2}}$$:

$$y'=\frac{(-2)(-2)-(-1)(-2)}{(-2)^{2}}=\frac{4-2}{4}=\frac12$$

Define $$A=N'D-ND'$$, so $$A' = N''D-ND''$$ (because the $$N'D'$$ terms cancel).
At $$\theta=\frac{\pi}{2}$$:

$$A=2,\;\;A'=4(-2)-(-1)(16)=-8+16=8$$

The second derivative of a quotient can be written as
$$y''=\frac{A'D-2AD'}{D^{3}}$$

Thus

$$y''=\frac{8(-2)-2(2)(-2)}{(-2)^{3}}=\frac{-16+8}{-8}=1$$

Finally, at $$\theta=\frac{\pi}{2}$$:

$$y''+y'+y = 1+\frac12+\frac12 = 2$$

The required value is 2, which matches Option C.