Question

Class -11

Quadratic Formula
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Answer 1

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

Given that,

$\rm \: \alpha , \beta ,\gamma \: \in \: R$

such that,

$\rm \: a { \alpha }^{2} + b \alpha + c = (sin\theta ) { \alpha }^{2} + (cos\theta ) \alpha - - - (1)$

$\rm \: a { \beta }^{2} + b \beta + c = (sin\theta ) { \beta }^{2} + (cos\theta ) \beta - - - (2)$

$\rm \: a { \gamma }^{2} + b \gamma + c = (sin\theta ) { \gamma }^{2} + (cos\theta ) \gamma - - - (3)$

From these 3 equations, we concluded that

$\rm \: \alpha, \beta , \gamma \: are \: the \: roots \: of \: the \: equation \: \:$

$\rm \: {ax}^{2} + bx + x = {(sin\theta )}x^{2} + x(cos\theta )$

can be rewritten as

$\rm \: {ax}^{2} + bx + x - {(sin\theta )x}^{2} - x(cos\theta ) = 0$

$\rm \: (a - sin\theta ) {x}^{2} + (bcos\theta )x + c = 0$

Since, its a quadratic equation in x and having 3 roots.

So, its an identity and its possible only, when

$\rm \:a - sin\theta = 0, \: \: b - cos\theta = 0, \: \: c = 0$

$\rm\implies \:a = sin\theta$

$\rm\implies \:b = cos\theta$

$\rm\implies \:c = 0$

Now, Consider

$\rm \: \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} + 3ab + {5b}^{2} }$

So, on substituting the values of a and b, we get

$\rm \: =  \: \dfrac{ {sin}^{2}\theta + {cos}^{2}\theta }{ {sin}^{2}\theta + 3sin\theta \: cos\theta + {5cos}^{2}\theta }$

$\rm \: =  \: \dfrac{1}{ {sin}^{2}\theta + 3sin\theta \: cos\theta + {5cos}^{2}\theta }$

can be further rewritten as

$\rm \: =  \: \dfrac{2}{ 2{sin}^{2}\theta + 3(2sin\theta \: cos\theta) + {5(2cos}^{2}\theta)}$

$\rm \: =  \: \dfrac{2}{1 - cos2\theta + 3sin2\theta + 5(1 + cos2\theta )}$

$\rm \: =  \: \dfrac{2}{1 - cos2\theta + 3sin2\theta + 5 + 5cos2\theta }$

$\rm \: =  \: \dfrac{2}{6 + 3sin2\theta + 4cos2\theta }$

So,

$\rm\implies \: \: \dfrac{2}{6 + 3sin2\theta + 4cos2\theta } \: is \: maximum \: when$

$\rm \: 6 + 3sin2\theta + 4cos2\theta \: is \: minimum.$

We know,

$\rm \: asinx + bcosx + c \: \in \: [c - \sqrt{ {a}^{2} + {b}^{2} }, \: c + \sqrt{ {a}^{2} + {b}^{2} }]$

So, it implies,

$\rm \: 6 + 3sin2\theta + 4cos2\theta \in \: [6 - \sqrt{ {3}^{2} + {4}^{2} }, \: 6 + \sqrt{ {3}^{2} + {4}^{2} }]$

$\rm \: 6 + 3sin2\theta + 4cos2\theta \in \: [6 - \sqrt{ 9 + 16 }, \: 6 + \sqrt{ 9 + 16}]$

$\rm \: 6 + 3sin2\theta + 4cos2\theta \in \: [6 - 5, \: 6 + 5]$

$\rm \: 6 + 3sin2\theta + 4cos2\theta \in \: [1, \: 11]$

$\rm\implies \:Minimum \: value \: of \: 6 + 3sin2\theta + 4cos2\theta \: is \: 1$

Thus,

$\rm\implies \:Maximum \: value \: of \: \dfrac{2}{6 + 3sin2\theta + 4cos2\theta } \: is \: 2$

Hence,

$\rm \: \rm\implies \:Maximum \: value \: of \: \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} + 3ab + {5b}^{2} } \: is \: 2$

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

Given that,

$\rm \: \vec{V_1} = a\hat{i} + b\hat{j} + c\hat{k}$

On substituting the values of a, b, c, we get

$\rm \: \vec{V_1} = sin\theta \hat{i} + cos\theta \hat{j} + 0\hat{k}$

Now, Also given that,

$\rm \: \vec{V_2} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$

Now, we know angle between two vectors is given by

$\rm \: cosx = \dfrac{\vec{V_1}.\vec{V_2}}{ |\vec{V_1}| \: |\vec{V_2}| }$

So, on substituting the values, we get

$\rm \: cos\dfrac{\pi}{3} = \dfrac{sin\theta + cos\theta + 0}{ \sqrt{ {sin}^{2}\theta + {cos}^{2}\theta } \times \sqrt{1 + 1 + 2} }$

$\rm \: \dfrac{1}{2} = \dfrac{sin\theta + cos\theta }{2}$

$\rm \: sin\theta + cos\theta = 1$

On squaring both sides, we get

$\rm \: {sin}^{2} \theta + {cos}^{2} \theta + 2sin\theta \: cos\theta \: = \: 1$

$\rm \: 1+ 2sin\theta \: cos\theta \: = \: 1$

$\rm \: sin2\theta \: = \: 0$

$\rm\implies \:2\theta = 0,\pi,2\pi,3\pi,4\pi$

$\rm\implies \:\theta = 0, \dfrac{\pi}{2} ,\pi, \dfrac{3\pi}{2},2\pi$

As,

$\rm \:\theta = \pi, \dfrac{3\pi}{2} \: donot \: satisfy \: sin\theta + cos\theta = 1$

So,

$\rm\implies \:\theta = 0, \: \dfrac{\pi}{2} , \: 2\pi \: \in \: [0,2\pi]$