Question

If
$y = \frac{2}{ \sin(x) + \sqrt{3} \cos(x) }$
then the minimum value of y is
(1) 1
(2 2
(3)
$\frac{1}{ \sqrt{3} + 1}$
(4)1/2
with explanation.
​

Answer 1

Basic Concept Used :-

↝ HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

• Differentiate the given function f(x).

• let f'(x) = 0 and find critical numbers.

• Then, find the second derivative of f(x) i.e. f''(x).

• Apply those critical numbers in the second derivative.

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{{\bf \: \dfrac{d}{dx}secx = secxtanx}}$

$\boxed{{\bf \: \dfrac{d}{dx}x = 1}}$

$\boxed{{\bf \: \dfrac{d}{dx}k = 0}}$

$\boxed{{\bf \: \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{{\bf \: \dfrac{d}{dx}u.v = v \: \dfrac{d}{dx} \: u \: + \: u \: \dfrac{d}{dx} \: v}}$

$\boxed{{\bf \: cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{{\bf \: cos\dfrac{\pi}{6} = \dfrac{ \sqrt{3} }{2} }}$

$\boxed{{\bf \: sin\dfrac{\pi}{6} = \dfrac{1}{2}}}$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:y = \dfrac{2}{sinx + \sqrt{3}cosx }$

$\rm :\longmapsto\:y \: = \: \dfrac{1}{\dfrac{1}{2}sinx + \dfrac{ \sqrt{3} }{2}cosx}$

$\rm :\longmapsto\:y = \dfrac{1}{sin\dfrac{\pi}{6}sinx + cos \dfrac{\pi}{6} cosx}$

$\rm :\longmapsto\:y = \dfrac{1}{cos\bigg( x - \dfrac{\pi}{6} \bigg) }$

$\rm :\longmapsto\:y = sec\bigg( x - \dfrac{\pi}{6} \bigg) - - (1)$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}sec\bigg( x - \dfrac{\pi}{6} \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = sec\bigg( x - \dfrac{\pi}{6} \bigg)tan\bigg( x - \dfrac{\pi}{6} \bigg) - - (2)$

For maxima or minima,

$\rm :\longmapsto\:Put \: \dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:sec\bigg( x - \dfrac{\pi}{6} \bigg)tan\bigg( x - \dfrac{\pi}{6} \bigg) = 0$

$\rm :\longmapsto\:tan\bigg( x - \dfrac{\pi}{6} \bigg) = 0 \: \: as \: sec\bigg( x - \dfrac{\pi}{6} \bigg) \ne \: 0$

$\rm :\longmapsto\:x - \dfrac{\pi}{6} = 0$

$\bf\implies \:x = \dfrac{\pi}{6}$

Now, On differentiating equation (2) w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2}y}{ {dx}^{2}} = sec\bigg( x - \dfrac{\pi}{6} \bigg)\dfrac{d}{dx}tan\bigg( x - \dfrac{\pi}{6} \bigg) + tan\bigg( x - \dfrac{\pi}{6} \bigg)\dfrac{d}{dx}sec\bigg( x - \dfrac{\pi}{6} \bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y}{ {dx}^{2}} = {sec}^{3}\bigg( x - \dfrac{\pi}{6} \bigg) + sec\bigg( x - \dfrac{\pi}{6} \bigg) {tan}^{2}\bigg( x - \dfrac{\pi}{6} \bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y}{ {dx}^{2}} \: at \: \bigg( x = \dfrac{\pi}{6} \bigg)= {sec}^{3}0 + sec0 \times {tan}^{2}0$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y}{ {dx}^{2}} \: at \: \bigg( x = \dfrac{\pi}{6} \bigg)=1 + 1 \times 0$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y}{ {dx}^{2}} \: at \: \bigg( x = \dfrac{\pi}{6} \bigg)=1 > 0$

$\bf\implies \:y \: is \: minimum \: at \: x \: = \: \dfrac{\pi}{6}$

And minimum value is

$\rm :\longmapsto\:y = sec\bigg( \dfrac{\pi}{6} - \dfrac{\pi}{6} \bigg)$

$\rm :\longmapsto\:y = sec0$

$\bf\implies \:y = 1$

$\overbrace{ \underline { \boxed { \rm \therefore \: The \: minimum \: value \: of \:\dfrac{2}{sinx + \sqrt{3}cosx } \: is \:1}}}$

$\large{\boxed{\boxed{\bf{Option \: (1) \: is \: correct}}}}$