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Mathematics Question on Differential Calculus

If y(θ)=2cos⁡θ+cos⁡2θcos⁡3θ+4cos⁡2θ+5cos⁡θ+2y(\theta) = \frac{2\cos\theta + \cos2\theta}{\cos3\theta + 4\cos2\theta + 5\cos\theta + 2}, then at θ=π2,y′′+y′+y\theta = \frac{\pi}{2}, y'' + y' + y is equal to:

A

32\frac{3}{2}

B

1

C

12\frac{1}{2}

D

2

Answer

2

Explanation

Solution

Given:

y(θ)=2cos⁡θ+2cos⁡2θ−14cos⁡3θ+8cos⁡2θ+5cos⁡θ+2.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(θ)=12⋅11+cos⁡θ.y(\theta) = \frac{1}{2} \cdot \frac{1}{1 + \cos \theta}.

Step 1: Evaluate y(θ)y(\theta) at θ=π2\theta = \frac{\pi}{2}
Substitute θ=π2\theta = \frac{\pi}{2}: y(π2)=12⋅11+cos⁡(π2)=12⋅11+0=12.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′(θ)y'(\theta)
Differentiate y(θ)y(\theta) using the chain rule: y′(θ)=12(−1(1+cos⁡θ)2)⋅(−sin⁡θ).y'(\theta) = \frac{1}{2} \left(-\frac{1}{(1 + \cos \theta)^2}\right) \cdot (-\sin \theta). Evaluating at θ=π2\theta = \frac{\pi}{2}: y′(π2)=12⋅1(1+0)2=12.y'\left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot \frac{1}{(1 + 0)^2} = \frac{1}{2}.

Step 3: Second Derivative y′′(θ)y''(\theta)
Differentiate y′(θ)y'(\theta): y′′(θ)=12(cos⁡θ(1+cos⁡θ)2−sin⁡θ⋅2(1+cos⁡θ)⋅(−sin⁡θ)(1+cos⁡θ)4).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 θ=π2\theta = \frac{\pi}{2}: y′′(π2)=1.y''\left(\frac{\pi}{2}\right) = 1.

Step 4: Summing the Values
y′′(π2)+y′(π2)+y(π2)=1+12+12=2.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).