Question

If $x=\frac{4\alpha }{1+{\alpha }^2}$ and $y=\frac{2-2{\alpha }^2}{1+{\alpha }^2},$ where $\alpha$ is a real parameter, then $x^2-xy+y^2$ lines between

• A. $[2,6]$
• B. $[2,4]$
• C. $[4,6]$
• D. $Noneofthese$

Answer 1

Answer: (A)

$[2,6]$

Given $x=\frac{4\alpha }{1+{\alpha }^2}$ and $y=\frac{2-2{\alpha }^2}{1+{\alpha }^2}$

Let $\alpha =tan\theta$ than $x=\frac{2.2tan\theta }{1+ta{n}^2\theta }=2sin2\theta$

$y=\frac{2(1-{\alpha }^2)}{1+{\alpha }^2}=2\left[\frac{1+ta{n}^2\theta }{1+ta{n}^2\theta }\right]=2cos\theta$

$\therefore x^2+xy+y^2=(2sin2\theta {)}^2-(2sin2\theta \left)\right(2cos2\theta )+(2cos2\theta )$

$=4si{n}^2\theta +4co{s}^{2}2\theta -4sin2\theta cos2\theta$

$=4(si{n}^2\theta +co{s}^2\theta )-2sin4\theta$

$=4-2sin4\theta$

$1\le sin4\theta \le 1$ it lies b/w

$\therefore 4-2$ to $4+2$ $[2,6]$