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) \[(-1,1]\]

Correct Answer: B

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 \[a\in [1,\infty )\]