Question

please solve this question​

Answer 1

QUESTION:

$\sf If \ y =\dfrac{2}{sin \ \theta+\sqrt{3}\ cos \ \theta}\ ,then\ the\ minimum$

$\sf value\ of \ y\ is$

ANSWER:

• The minimum value of y = 1

GIVEN:

$\sf y =\dfrac{2}{sin \ \theta+\sqrt{3}\ cos \ \theta}$

TO FIND:

• The minimum value of y.

EXPLANATION:

$\sf y({sin \ \theta+\sqrt{3}\ cos \ \theta})=2$

Differentiate w.r.t x

$\boxed{\bold {\large{ \dfrac{d}{dx}uv = uv' + vu'}}}$

$\boxed{\bold {\large{ \dfrac{d}{dx}k = 0}}}$

$\sf y({cos \ \theta+\sqrt{3}\ (- sin \ \theta})) \\ \sf + \dfrac{dy}{dx}({sin \ \theta+\sqrt{3}\ cos \ \theta}) =0$

$\sf sin \ \theta+\sqrt{3}\ cos \ \theta=\dfrac{2}{y}$

$\sf \dfrac{dy}{dx} \left( \dfrac{2}{y} \right ) = - y({cos \ \theta - \sqrt{3}\ sin \ \theta})$

$\sf \dfrac{dy}{dx} = - \dfrac{ y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta})}{2}$

Put dy/dx = 0

$\sf 0 = - \dfrac{ y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta})}{2}$

$\sf 0 = cos \ \theta - \sqrt{3}\ sin \ \theta$

$\sf \sqrt{3}\ sin \ \theta = cos \ \theta$

$\sf \dfrac{sin \ \theta}{ \cos \theta} = \dfrac{1}{ \sqrt{3} }$

$\sf Tan \ \theta = \dfrac{1}{ \sqrt{3} }$

$\sf \theta ={ 30}^{ \circ}$

$\sf \dfrac{dy}{dx} = - \dfrac{ y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta})}{2}$

Differentiate w.r.t x

$\sf \dfrac{d ^{2} y}{dx^{2} } = - \dfrac{ 1}{2} \dfrac{d}{dx} \bigg(y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta} ) \bigg)$

$\sf Let \ \bigg(y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta} ) \bigg) \ \ be \ z$

$\sf \dfrac{d}{dx} \bigg(y ^{2} ({cos \ \theta - \sqrt{3}\ sin \ \theta} ) \bigg) = \dfrac{dz}{dx}$

$\sf \dfrac{dz}{dx} =2y({cos \theta - \sqrt{3}\ sin \theta} ) \\ \sf +{y}^{2} ( - sin \theta - \sqrt{3} cos \theta)$

$\sf \dfrac{d ^{2} y}{dx^{2} } = - \dfrac{ 1}{2}(2y({cos \theta - \sqrt{3}\ sin \theta} ) )\\ \sf + {y}^{2} ( - sin \theta - \sqrt{3} cos \theta)$

Take 2y(cos θ - √3 sin θ)

Substitute θ = 30°

cos 30° = √3 / 2

sin 30° = 1 / 2

2y(cos θ - √3 sin θ) = 2y(√3 / 2 - √3 / 2)

2y(cos θ - √3 sin θ) = 2y(0)

2y(cos θ - √3 sin θ) = 0

$\sf \dfrac{d ^{2} y}{dx^{2} } = - \dfrac{ 1}{2}( {y}^{2} ( - sin \theta - \sqrt{3} cos \theta))$

$\sf \dfrac{d ^{2} y}{dx^{2} } = \dfrac{ 1}{2}( {y}^{2} \ sin \ \theta + {y}^{2} \ \sqrt{3} \ cos \ \theta)$

Sin θ and cos θ are positive when θ = 30°

$\sf \therefore \ \dfrac{d ^{2} y}{dx^{2} } > 0$

Hence y is minimum at θ = 30°

$\sf y =\dfrac{2}{sin \ \theta+\sqrt{3}\ cos \ \theta}$

Substitute θ = 30°

$\sf y =\dfrac{2}{sin \ {30}^{ \circ} +\sqrt{3}\ cos \ {30}^{ \circ}}$

$\sf y =\dfrac{2}{ \dfrac{1}{2} +\sqrt{3} \left(\dfrac{ \sqrt{3} }{2}\right)}$

$\sf y =\dfrac{2}{ \dfrac{1 + 3}{2} }$

$\sf y =\dfrac{2}{ \dfrac{4}{2} }$

$\sf y = \dfrac{2}{2}$

$\sf y = 1$

Hence the minimum value of y = 1