Q. The range of $\alpha$, which the point $(\alpha,\alpha)$ lies inside the region bounded by the curves $y = \sqrt{1 - x^{2}}$ and $x + y = 1$ is
A) $\frac{1}{2} < \alpha < \frac{1}{\sqrt{2}}$
B) $\frac{1}{2} < \alpha < \frac{1}{3}$
C) $\frac{1}{3} < \alpha < \frac{1}{\sqrt{3}}$
D) $\frac{1}{4} < \alpha < \frac{1}{2}$
Explanation
(a)
The point should lies on the opposite side of the origin of the line $x + y - 1 = 0$
[image:11_image59.emf]
Then, $\alpha + \alpha - 1 > 0$
$\Rightarrow 2\alpha > 1\ \Rightarrow \ \alpha > \frac{1}{2}$ …(i)
Also, $(\alpha^{2} + \alpha^{2} < 1$
$\Rightarrow \ \ \left( - \frac{1}{\sqrt{2}} \right) < \alpha < \left( \frac{1}{\sqrt{2}} \right)$
From relation (i) and (ii), we get
$\frac{1}{2} < \alpha < \frac{1}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.