Question

If $x=\frac{4\lambda }{1+{\lambda }^2}$ and $y=\frac{2-2{\lambda }^2}{1+{\lambda }^2}$ where $\lambda$ is a real parameter and $x^2–xy+y^2$ lies between $[a,b]$ then $(a+b)$ is

• A. 8
• B. 10
• C. 13
• D. 25

Answer 1

The correct option is A

8

Let $\lambda =\tan \theta$
$\Rightarrow x=2\sin 2\theta$ and $y=2\cos 2\theta$ [$\because \sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$ and $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$]
Given, $f=x^2-xy+y^2$
$=4-4\sin 2\theta \cos 2\theta =4-2\sin 4\theta$

We know that $-1\le \sin 4\theta \le 1$

$\Rightarrow 2\ge -2\sin 4\theta \ge -2$

$\Rightarrow 6\ge 4-2\sin 4\theta \ge 2$
$\therefore f$ lies between 2 and 6 or $fϵ[2,6]$
$\therefore a=2$ and $b=6\Rightarrow a+b=8$