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 $a$ in the interval are

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

Answer 1

Answer: (A)

$(1,\infty )$

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

solving these equations simultaneously,

we get,

$x=\frac{\left|a\right|+1}{1+a},y=\frac{a\left|a\right|-1}{1+a}$

As both are in the $1^{st}$ Quadrant both should be greater than $0$,

$1+a>0$ and $a\left|a\right|-1>0$

$a>-1$ and $a|a|>1$

After solving this, we get

$a>1,$ i.e $aϵ(1,\infty )$