• # question_answer 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:     AIEEE  Solved  Paper (Held On 11 May  2011) A) $(0,\infty )$ B) $[1,\infty )$ C) $(-1,\infty )$ D) $(-1,1]$

Solution :

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

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