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Question Number 146876 by gsk2684 last updated on 16/Jul/21

i f t h e m a x i m u m v a l u e o f

4 \sin^{2} x + 3 \cos^{2} x + \sin \frac{x}{2} + \cos \frac{x}{2} + 3

i s a + \sqrt{b} t h e n f i n d a + b

Answered by liberty last updated on 16/Jul/21

f (x) = 4 \sin^{2} x + 3 - 3 \sin^{2} x + 3 + \sin (\frac{x}{2}) + \cos (\frac{x}{2})

f (x) = \sin^{2} x + \sin (\frac{x}{2}) + \cos (\frac{x}{2}) + 6

f (x) = \sin^{2} x + \sin (\frac{x}{2}) + \cos (\frac{x}{2}) + 6

f^{'} (x) = \sin 2 x + \frac{1}{2} \cos (\frac{x}{2}) - \frac{1}{2} \sin (\frac{x}{2}) = 0

\sin 2 x = \frac{1}{2} \sin (\frac{x}{2}) - \frac{1}{2} \cos (\frac{x}{2})

2 \sin 2 x - \sin (\frac{x}{2}) + \cos (\frac{x}{2}) = 0

l e t \frac{x}{2} = u \to x = 2 u

2 \sin 4 u + (\cos u - \sin u) = 0

4 \sin 2 u (\cos^{2} u - \sin^{2} u) + (\cos u - \sin u) = 0

(\cos u - \sin u) {4 \sin 2 u (\cos u + \sin u) + 1} = 0

(∙) \cos u - \sin u = 0

\cos u = \sin u \Rightarrow \tan (\frac{x}{2}) = 1

x = \frac{π}{2}

f (\frac{π}{2}) = 1 + 6 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 7 + \sqrt{2}

f_{m a x} = 7 + \sqrt{2} \approx 8.414213

t h e n {\begin{cases} a = 7 \\ b = 2 \end{cases} \Rightarrow a + b = 9

Commented by gsk2684 last updated on 16/Jul/21

t h a n k y o u

Commented by liberty last updated on 16/Jul/21

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