If = 2 cos + cos 2 cos 3 + 4 cos 2 + 5 cos + 2 , then at = 2 , y ' ' + y ' + y is equal to [2024]

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Solution

Q.
$If (θ) = \frac{2 \cos θ + \cos 2 θ}{\cos 3 θ + 4 \cos 2 θ + 5 \cos θ + 2}, then at θ = \frac{π}{2}, y'' + y' + y is equal to$ [2024]

1 1

2 $\frac{3}{2}$

3 2

4 $\frac{1}{2}$

Ans.
(3)

We have, $y = \frac{2 \cos θ + \cos 2 θ}{\cos 3 θ + 4 \cos 2 θ + 5 \cos θ + 2}$

$= \frac{2 \cos θ + 2 \cos^{2} θ - 1}{4 \cos^{3} θ - 3 \cos θ + 8 \cos^{2} θ - 4 + 5 \cos θ + 2}$

$= \frac{2 \cos^{2} θ + 2 \cos θ - 1}{4 \cos^{3} θ + 8 \cos^{2} θ + 2 \cos θ - 2}$

$= \frac{2 \cos^{2} θ + 2 \cos θ - 1}{2 \cos θ (2 \cos^{2} θ + 2 \cos θ - 1) + 4 \cos^{2} θ + 4 \cos θ - 2}$

$= \frac{2 \cos^{2} θ + 2 \cos θ - 1}{2 \cos θ (2 \cos^{2} θ + 2 \cos θ - 1) + 2 (2 \cos^{2} θ + 2 \cos θ - 1)}$

$= \frac{1}{2 \cos θ + 2}$

$= \frac{1}{2} (\frac{1}{\cos θ + 1})$

$y' (θ) = \frac{1}{2} [\frac{\sin θ}{{(1 + \cos θ)}^{2}}]$

$y'' (θ) = \frac{1}{2} [\frac{\cos θ {(1 + \cos θ)}^{2} + \sin^{2} θ \cdot 2 (1 + \cos θ)}{{(1 + \cos θ)}^{4}}]$

$= \frac{1}{2} [\frac{\cos θ (1 + \cos θ) + 2 \sin^{2} θ}{{(1 + \cos θ)}^{3}}]$

at $θ = \frac{π}{2}, y' = \frac{1}{2}, y'' = 1$ and $y = \frac{1}{2}$

So, $y'' + y' + y = 1 + \frac{1}{2} + \frac{1}{2} = 2$

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