If $a > b$ , minimum value of $a {sin}^{2} x + b {cos}^{2} x$

a

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b

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a + b

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\sqrt{(a^{2} + b^{2})}

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Solution

The correct option is B $b$
$f (x) = a {sin}^{2} x + b {cos}^{2} x$
$f^{'} (x) = (a - b) sin 2 x$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow sin 2 x = 0$
$\Rightarrow x = 0, \frac{π}{2}$

$f^{''} (x) = 2 (a - b) cos 2 x$
$f^{''} (0) = 2 (a - b)$
Since, $a > b,$
So, $f^{''} (0) > 0$
So, minimum value of f(x) is at $x = 0$
$f (0) = b$

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