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Q.

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

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a

a

b

b

c

a+b

d

$\sqrt{a^{2} + b^{2}}$

answer is B.

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Detailed Solution

$a > b f (x) = {asin}^{2} x + {bcos}^{2} x = {asin}^{2} x + b (1 - \sin^{2} x) = b + (a - b) \sin^{2} x$

Minimum value of

$\sin^{2} x = 0 ∴ \min of f (x) = b + 0 = b$

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