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:

• 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 1

The Correct Option is B

Given the lines \(3x - y + 1 = 0\) and \(x + 2y - 5 = 0\), we need to find the region \(R\) that is bounded by these lines and contains the origin.

Line Equations Analysis:
For the line \(3x - y + 1 = 0\), rearranging gives \(y = 3x + 1\).  
For the line \(x + 2y - 5 = 0\), rearranging gives \(y = \frac{5 - x}{2}\).  

The region \(R\) is bounded by these lines such that it includes the origin \((0, 0)\).

Condition for Points \((a^2, a + 1)\) to Lie in \(R\):
The point \((a^2, a + 1)\) lies in \(R\) if it satisfies the inequalities:  
\(3a^2 + 1 < a + 1 \quad \text{and} \quad a + 1 < \frac{5 - a^2}{2}.\)

Simplifying the first inequality:  
\(3a^2 + 1 < a + 1 \implies 3a^2 - a < 0 \implies a(3a - 1) < 0.\)
This gives the interval \(-\frac{1}{3} < a < 0\).

Simplifying the second inequality:  
\(a + 1 < \frac{5 - a^2}{2} \implies 2a + 2 < 5 - a^2 \implies a^2 + 2a - 3 > 0.\)
Factoring gives:  
\((a - 1)(a + 3) > 0.\)
This gives the intervals \(a < -3\) or \(a > 1\).

Combining the Intervals:
The valid values of \(a\) are the intersection of \(-\frac{1}{3} < a < 0\) with \(a < -3\) or \(a > 1\), which results in:  
\((-3, 0) \cup \left(-\frac{1}{3}, 1\right).\)

Thus, the correct answer is : \( (-3, 0) \cup \left(\frac{1}{3}, 1\right) \)