Question

The value of x for the minimum value of √3 cos +sin is :
180 degree
270 degree
210 degree
0 degree​

Answer 1

Answer:

270 degree.............

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The value of x for which √3 cos x +sin x is minimum

CALCULATION

$\sf{Let \: \: f(x) = \sqrt{3} \cos x + \sin x \: \: }$

Differentiating both sides with respect to x two times

$\sf{ {f} \: {'}(x) = - \sqrt{3} \sin x + \cos x \: \: }$

$\sf{ {f} \: {''}(x) = - ( \sqrt{3} \cos x + \sin x \:) \: }$

For extremum value of f(x) we have

$\sf{ {f} \: {'}(x) = 0\: }$

$\implies \: \sf{ - \sqrt{3} \sin x + \cos x \: = 0 \: }$

$\implies \: \sf{ \sqrt{3} \sin x = \cos x \: \: }$

$\implies \: \sf{ \cot x \: = \sqrt{3} \: }$

$\implies \: \sf{ x = \: {30}^{ \circ} , {210}^{ \circ}\: }$

Now

$\sf{ {f} \: {''}( {30}^{ \circ} ) = - ( \sqrt{3} \cos {30}^{ \circ}+ \sin {30}^{ \circ} \:) = - 2 < 0}$

So f(x) has a maximum value at x = 30°

Again

$\sf{ {f} \: {''}( {210}^{ \circ} ) = - ( \sqrt{3} \cos {210}^{ \circ}+ \sin {210}^{ \circ} \:) = 2 > 0}$

So f(x) has a minimum value at x = 210°

The minimum value is

$\sf{ {f}( {210}^{ \circ} ) = ( \sqrt{3} \cos {210}^{ \circ}+ \sin {210}^{ \circ} \:) = - 2 \: }$

RESULT

For √3 cos x +sin x the minimum value occurs at x = 210° and the minimum value is - 2