Question

If sinα and cosα are the roots of x²+ax+b=0, then for any real number α, the range of a is __________ and the range of b is ___________​

Answer 1

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

Given that,

$\red{\rm :\longmapsto\:sin \alpha \: and \: cos \alpha \: are \: roots \: of \: {x}^{2} + ax + b = 0}$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\:sin \alpha + cos \alpha = - \dfrac{a}{1}$

$\bf\implies \:a = - (sin \alpha + cos \alpha)$

Now, To find the range of a, we have to find the range of - ( sinα + cosα )

Now, Consider,

$\rm :\longmapsto\: - (sin \alpha + cos \alpha)$

can be rewritten as

$\rm \: = \: \: - \sqrt{2}\bigg(\dfrac{1 }{ \sqrt{2} }sin \alpha + \dfrac{1}{ \sqrt{2} } cos \alpha \bigg)$

$\rm \: = \: \: - \sqrt{2}\bigg(cos\dfrac{\pi }{ 4 }sin \alpha + sin\dfrac{\pi }{ 4 } cos \alpha \bigg)$

$\rm \: = \: \: - \sqrt{2} sin\bigg(x + \dfrac{\pi }{ 4 } \bigg)$

So, it means

$\rm :\longmapsto\: \: - (sin \alpha + cos \alpha)= \: \: - \sqrt{2} sin\bigg(x + \dfrac{\pi }{ 4 } \bigg)$

$\bf\implies \:a \: = \: - \sqrt{2} sin\bigg(x + \dfrac{\pi }{ 4 } \bigg)$

Now, we know that,

$\rm :\longmapsto\: - 1 \leqslant sin\bigg(x + \dfrac{\pi }{ 4 } \bigg) \leqslant 1$

$\rm :\longmapsto\: - \sqrt{2} \leqslant \: - \sqrt{2} sin\bigg(x + \dfrac{\pi }{ 4 } \bigg) \: \leqslant \sqrt{2}$

$\bf :\longmapsto\: - \sqrt{2} \leqslant \: a\: \leqslant \sqrt{2}$

So, this is the range of a.

Now, we also know that

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\:sin \alpha \: cos \alpha = \dfrac{b}{1}$

$\rm :\implies\:b \: = \: sin \alpha \: cos \alpha$

can be rewritten as

$\rm :\implies\:b \: = \: \dfrac{1}{2} \times (2sin \alpha \: cos \alpha)$

$\rm :\implies\:b \: = \: \dfrac{1}{2} \times ( \: sin 2\alpha \: )$

Now, we know,

$\rm :\longmapsto\: - 1 \leqslant sin2 \alpha \leqslant 1$

$\rm :\longmapsto\: - \dfrac{1}{2} \leqslant \dfrac{1}{2}sin2 \alpha \leqslant \dfrac{1}{2}$

$\rm :\longmapsto\: - \dfrac{1}{2} \leqslant b \leqslant \dfrac{1}{2}$

This, is the required range of b.

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$