Animated Solution for Mathematics - Differentiation: If $y(\theta )=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}$, then at $\theta =\frac{\pi }{2}$, $y^{\prime \prime }+y^{\prime }+y$ is equal to :

Welcome, warrior of mathematics. Today, we stand before a problem that looks like a nightmare. You see a fraction, you see multiple angles like $2\theta$ and $3\theta$, and your instinct might be to panic and reach for the quotient rule.

But stop. Take a breath. In the world of JEE Advanced, the most intimidating problems are often the ones that hide the simplest truths.

We are given the function:

We need to find $y^{\prime \prime }+y^{\prime }+y$ at $\theta =\frac{\pi }{2}$. If you dive into differentiation now, you will drown in algebra. Let us instead use the scalpel of trigonometry to dissect this beast.

We must strip away the complexity. Our goal is to express everything in terms of $\cos \theta$.

We recall our trusty identities:

Let us apply these to the numerator first. We have $2\cos \theta +\cos 2\theta$. Substituting our identity, we get $2\cos \theta +(2{\cos }^2\theta -1)$, which rearranges beautifully to $2{\cos }^2\theta +2\cos \theta -1$. Keep this safe; it is the key to our salvation.

Now, consider the denominator: $\cos 3\theta +4\cos 2\theta +5\cos \theta +2$. Substituting our identities, we get:

Expanding this, we find $4{\cos }^3\theta -3\cos \theta +8{\cos }^2\theta -4+5\cos \theta +2$. Combining like terms, we arrive at $4{\cos }^3\theta +8{\cos }^2\theta +2\cos \theta -2$. If we factor out a $2$, we get $2(2{\cos }^3\theta +4{\cos }^2\theta +\cos \theta -1)$.

Here is the moment of truth. We have a cubic polynomial. By testing $\cos \theta =-1$, we see the expression vanishes, meaning $(\cos \theta +1)$ is a factor.

Through polynomial division, we find the other factor is exactly our numerator: $(2{\cos }^2\theta +2\cos \theta -1)$. The entire expression for $y(\theta )$ collapses into:

The quadratic terms cancel out! We are left with the incredibly simple expression:

Now, the calculus becomes a joy. We need $y(\theta )$, $y^{\prime }(\theta )$, and $y^{\prime \prime }(\theta )$ at $\theta =\frac{\pi }{2}$.

First, $y(\frac{\pi }{2})=\frac{1}{2(0+1)}=\frac{1}{2}$.

Next, for $y^{\prime }(\theta )$, we rewrite $y(\theta )=\frac{1}{2}(\cos \theta +1{)}^{-1}$. Using the chain rule:

At $\theta =\frac{\pi }{2}$, $\sin (\frac{\pi }{2})=1$ and $\cos (\frac{\pi }{2})=0$, so $y^{\prime }(\frac{\pi }{2})=\frac{1}{2(1{)}^2}=\frac{1}{2}$.

Finally, for $y^{\prime \prime }(\theta )$, we apply the quotient rule to $y^{\prime }(\theta )$. After careful differentiation and substituting $\theta =\frac{\pi }{2}$, we find $y^{\prime \prime }(\frac{\pi }{2})=1$.

Adding them all together: $1+\frac{1}{2}+\frac{1}{2}=2$.

We have conquered the monster. The final answer is 2. Remember, in JEE, the path to the answer is often hidden in plain sight—you just need the courage to simplify.