JEE Mains · Maths · STD 11 - 5. linear inequalities
Let \(R\) be the interior region between the lines \(3 x-y+1=0\) and \(x+2 y-5=0\) containing the origin. The set of all values of a, for which the points \(\left(a^2, a+1\right)\) lie in \(R\), is :
- A \((-3,-1) \cup\left(-\frac{1}{3}, 1\right)\)
- B \((-3,0) \cup\left(\frac{1}{3}, 1\right)\)
- C \((-3,0) \cup\left(\frac{2}{3}, 1\right)\)
- D \((-3,-1) \cup\left(\frac{1}{3}, 1\right)\)
Answer & Solution
Correct Answer
(B) \((-3,0) \cup\left(\frac{1}{3}, 1\right)\)
Step-by-step Solution
Detailed explanation
\( \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right)\) \( \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0\) Origin and \(P\) lies same side w.r.t. \(L_1\) \( \Rightarrow L_1(0) \cdot L_1(P)>0 \) \(\therefore 3\left(a^2\right)-(a+1)+1>0\) \( \Rightarrow 3 a^2-a>0 \)…
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