Math, asked by aditi2369, 3 months ago

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

what is the answer? with solution?

Answers

Answered by hotcupid16
138

\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}}}}

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