Straight Lines
Region Between Two Lines — Values of Parameter
nta_pyq_2024_jan
Grade 11
Question:
Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$, for which the points $(a^2,a+1)$ lie in $R$, is:
$(-3,-1)\cup\left(-\dfrac{1}{3},1\right)$
$(-3,0)\cup\left(\dfrac{1}{3},1\right)$
$(-3,0)\cup\left(\dfrac{2}{3},1\right)$
$(-3,-1)\cup\left(\dfrac{1}{3},1\right)$
Step-by-Step Solution
Key Concept: Origin satisfies $L_1(0,0)=1>0$ and $L_2(0,0)=-5<0$. For point $(a^2,a+1)$ to be in same region: $3a^2-(a+1)+1>0$ and $a^2+2(a+1)-5<0$. Solve both inequalities and take intersection.
$a\in(-3,0)\cup\left(\frac{1}{3},1\right)$.
Correct Answer: 2