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Trigonometric Equations

If a = 1, b =...

Question

If $a = 1, b = \sqrt{3}, c = 4, f (x) = cos x, g (x) = sin x$ and $h (x) = 1$ then minimum value of F(x) occurs at

A

\frac{- π}{3}

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B

\frac{- 2 π}{3}

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C

\frac{2 π}{3}

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D

\frac{π}{3}

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Solution

The correct option is B $\frac{- 2 π}{3}$
$F (x) = a f (x) + b g (x) + c h (x)$
$= cos x + \sqrt{3} sin x + 1$
at $x = - \frac{2 π}{3}$ funtion has minimum value.

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Similar questions

f (x) = \frac{1}{(cos x - 3)^{2} + (sin x + 4)^{2}}

g (x) = \sqrt{3} sin x + cos x

. Then which of the following is/are true ?

f (x) = x^{2} + \frac{1}{x^{2}}

g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}

h (x) = \frac{f (x)}{g (x)}

. Then the local minimum value of

h (x)

f (x) = x^{2} - \frac{1}{x^{2}}

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

f (x) = x^{2} + \frac{1}{x^{2}}

g (x) = x - \frac{1}{x},

x \in R - {- 1, 0, 1} .

h (x) = \frac{f (x)}{g (x)},

then the local minimum value of

h (x)

f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}, and h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

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