Sum of the maximum and minimum values of $12 {cos}^{2} x - 6 sin x cos x + 2 {sin}^{2} x$ is

0

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7

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14

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15

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Solution

The correct option is C $14$
$S = 12 {cos}^{2} x - 6 sin x cos x + 2 {sin}^{2} x$
$= 2 ({cos}^{2} x + {sin}^{2} x) + 10 {cos}^{2} x - 6 sin x cos x$
$= 2 + 10 {cos}^{2} x - 3 (2 sin x c o s x)$
$= 2 + 5 (2 {cos}^{2} x - 1) + 5 - 3 (2 sin x cos x)$
$= 2 + 5 cos 2 x + 5 - 3 (sin 2 x)$
$= 7 + 5 cos 2 x - 3 s i n 2 x$
$= 7 + (\frac{5 cos 2 x}{\sqrt{5^{3} + 3^{2}}} - \frac{3 sin 2 x}{\sqrt{5^{2} + 3^{2}}}) \sqrt{5^{2} + 3^{2}}$
$S = 7 + \sqrt{34} c o s α$
$S_{m i n} = 7 - \sqrt{34}$
$S_{m a x} = 7 + \sqrt{34}$
$S_{m i n} + S_{m a x} = 14$

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