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

The Correct Option is B

$x + y = |a|$ $ax - y = 1$ if $a > 0$ $x + y = a$ $ax - y = 1$ $x\left(1 + a\right) = 1 + a \,as\, x = 1$ $y = a - 1$ 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} $ $= \frac{-a-a^{2}-1+a}{1+a}> 0$ $-\left(\frac{a^{2}+1}{a+1}\right)> 0\quad\Rightarrow\quad \frac{a^{2}+1}{a+1} < 0$ from $\left(1\right)$ and $\left(2\right) a \in \left\{\phi\right\}$ So correct answer is a $[\in 1, \infty)$