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

Given:

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

Factoring the numerator and denominator, we get:

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

Step 1: Evaluate \(y(\theta)\) at \(\theta = \frac{\pi}{2}\)
Substitute \(\theta = \frac{\pi}{2}\): \[ y\left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot \frac{1}{1 + \cos\left(\frac{\pi}{2}\right)} = \frac{1}{2} \cdot \frac{1}{1 + 0} = \frac{1}{2}. \]

Step 2: First Derivative \(y'(\theta)\)
Differentiate \(y(\theta)\) using the chain rule: \[ y'(\theta) = \frac{1}{2} \left(-\frac{1}{(1 + \cos \theta)^2}\right) \cdot (-\sin \theta). \] 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: Second Derivative \(y''(\theta)\)
Differentiate \(y'(\theta)\): \[ y''(\theta) = \frac{1}{2} \left(\frac{\cos \theta (1 + \cos \theta)^2 - \sin \theta \cdot 2(1 + \cos \theta) \cdot (-\sin \theta)}{(1 + \cos \theta)^4}\right). \] Simplifying and evaluating at \(\theta = \frac{\pi}{2}\): \[ y''\left(\frac{\pi}{2}\right) = 1. \]

Step 4: Summing the Values
\[ y''\left(\frac{\pi}{2}\right) + y'\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right) = 1 + \frac{1}{2} + \frac{1}{2} = 2. \]

Therefore, the correct answer is Option (4).