Question

The lines $x+y=|a|$ and $ax-y=1$ intersect each other in the first quadrant. Then the set of all possible values of $a$ is the interval :

• A. $\left(0,\infty \right)$
• B. $[1,\infty )$
• C. $\left(-1,\infty \right)$
• D. $(-1,1]$

Answer 1

Answer: (B)

$x+y=|a|$
$ax-y=1$
if $a>0$
$x+y=a$
$ax-y=1$
$x\left(1+a\right)=1+aasx=1$
$y=a-1$
It is first quadrant
so $a-1\ge 0$
$a\ge 1$
$a\in [1,\infty )$
If $a<0$
$x+y=-a$
$ax-y=1$
$+$
$x\left(1+a\right)=1-a$
$x=\frac{1-a}{1+a}>0\Rightarrow \frac{a-1}{a+1}<0$
$y=-a-\frac{1-a}{1+a}$
$=\frac{-a-a^2-1+a}{1+a}>0$
$-\left(\frac{a^2+1}{a+1}\right)>0\Rightarrow \frac{a^2+1}{a+1}<0$
from $\left(1\right)$ and $\left(2\right)a\in \left\{\varphi \right\}$
So correct answer is a $[\in 1,\infty )$