Question

If $y(\theta) = \frac{2\cos\theta + \cos2\theta}{\cos3\theta + 4\cos2\theta + 5\cos\theta + 2}$, then at $\theta = \frac{\pi}{2}, y'' + y' + y$ is equal to:

• A. $\frac{3}{2}$
• B. 1
• C. $\frac{1}{2}$
• D. 2

Answer 1

The Correct Option is D

Provided:

\[ y(\theta) = \frac{2 \cos \theta + 2 \cos^2 \theta - 1}{4 \cos^3 \theta + 8 \cos^2 \theta + 5 \cos \theta + 2}. \]

After factoring the numerator and denominator, the expression simplifies to:

\[ y(\theta) = \frac{1}{2} \cdot \frac{1}{1 + \cos \theta}. \]

Step 1: Evaluate \(y(\theta)\) at \(\theta = \frac{\pi}{2}\)
Substituting \(\theta = \frac{\pi}{2}\) yields:

\[ y\left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot \frac{1}{1 + \cos\left(\frac{\pi}{2}\right)} = \frac{1}{2} \cdot \frac{1}{1 + 0} = \frac{1}{2}. \]

Step 2: Calculate the First Derivative \(y'(\theta)\)
Differentiating \(y(\theta)\) using the chain rule gives:

\[ y'(\theta) = \frac{1}{2} \left(-\frac{1}{(1 + \cos \theta)^2}\right) \cdot (-\sin \theta). \]Evaluating at \(\theta = \frac{\pi}{2}\):

\[ y'\left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot \frac{1}{(1 + 0)^2} = \frac{1}{2}. \]

Step 3: Calculate the Second Derivative \(y''(\theta)\)
Differentiating \(y'(\theta)\) results in:

\[ y''(\theta) = \frac{1}{2} \left(\frac{\cos \theta (1 + \cos \theta)^2 - \sin \theta \cdot 2(1 + \cos \theta) \cdot (-\sin \theta)}{(1 + \cos \theta)^4}\right). \]Simplifying and evaluating at \(\theta = \frac{\pi}{2}\):

\[ y''\left(\frac{\pi}{2}\right) = 1. \]

Step 4: Sum the Evaluated Derivatives and Function Value
The sum is:

\[ y''\left(\frac{\pi}{2}\right) + y'\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right) = 1 + \frac{1}{2} + \frac{1}{2} = 2. \]

The final result corresponds to Option (4).