Range ? Of which function!!

Algebra Level 4

$x=\frac { 4\lambda }{ 1+{ \lambda }^{ 2 } }$ $y=\frac { 2-2{ \lambda }^{ 2 } }{ 1+{ \lambda }^{ 2 } }$ $\lambda \in \Re$ If the range of ${ x }^{ 2 }+{ y }^{ 2 }-xy$ is [a,b] , find a+b.

4 8 0 6 5

1 solution

Chew-Seong Cheong
Apr 23, 2015

Let $\lambda = \tan{\frac{\theta}{2}}$ .

$\begin{cases} x = \dfrac {4\lambda}{1+\lambda^2} = \dfrac {4\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}} = 2\sin{\theta} \\ y = \dfrac {2-2\lambda^2}{1+\lambda^2} = \dfrac {2\left(1-\tan^2{\frac{\theta}{2}}\right)}{1+\tan^2{\frac{\theta}{2}}} = 2\cos{\theta} \end{cases}$

$\Rightarrow x^2+y^2-xy = 4\sin^2{\theta} + 4\cos^2{\theta} - 4\sin{\theta}\cos{\theta} = 4-2\sin{2\theta}$

Since $\space \sin{2\theta} \in [-1,1] \quad \Rightarrow x^2+y^2-xy \in [2,6] \quad \Rightarrow a + b = 2+6 = \boxed{8}$

same method sir :) @Chew-Seong Cheong

Madhukar Thalore - 6 years ago

Range ? Of which function!!

1 solution

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