Question

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

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

Answer 1

Answer: (B)

$[1,\infty )$

Given lines are :

$x+y=\left|a\right|$ and $ax-y=1$

Case $1:a>0$

$x+y=a-----\left(1\right)$

and $ax-y=1------\left(2\right)$

Adding $\left(1\right)+\left(2\right)$

$\Rightarrow x(1+a)=1+a$

$\Rightarrow x=1$

Hence, $y=a-1$

Since it is in first quadrant,$a-1\ge 0$

$\Rightarrow a\ge 1$

Case $2:a<0$

$x+y=-a$ and $ax-y=1$

Solving For $x,y$

$x=\frac{1-a}{1+a}>0$

$\Rightarrow \frac{a-1}{a+1}<0$

$\Rightarrow aϵ(-1,1)$

also ,$y=-a-\left(\frac{1-a}{1+a}\right)$ which should be $>0$

$\Rightarrow -\frac{a^2+1}{a+1}>0$

$\Rightarrow a<-1$

Combining both two cases, we get :

$a\ge 1$

$aϵ[1,\infty )$