Question

The range of $\alpha$, for 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. $\frac{1}{4}<\alpha <\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}$

The point should lies on the opposite side of the origin of the line $x+y-1=0$
Then, $\alpha +\alpha -1>0$
$\Rightarrow 2\alpha >1\Rightarrow \alpha >\frac{1}{2}.....\left(i\right)$
Also, $({\alpha }^2+{\alpha }^2)<1$
$\Rightarrow \left(\frac{-1}{\sqrt{2}}\right)<\alpha <\left(\frac{1}{\sqrt{2}}\right)....\left(ii\right)$
From Eqs. (i) and (ii), we get
$\frac{1}{2}<\alpha <\frac{1}{\sqrt{2}}$.