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

To determine the set of all values of \( a \) such that the points \( (a^2, a + 1) \) lie in the region \( R \), we first examine the region bounded by the lines \( 3x - y + 1 = 0 \) and \( x + 2y - 5 = 0 \) that contains the origin.

First, rewrite the equations of the lines: 

• \(3x - y + 1 = 0 \ \Rightarrow \ y = 3x + 1\)
• \(x + 2y - 5 = 0 \ \Rightarrow \ y = \frac{5-x}{2}\)

To find the region that contains the origin \((0, 0)\), substitute \( (0, 0) \) into the inequalities obtained from these lines:

• For \(3x - y + 1 = 0\), the inequality is \(3x - y + 1 > 0\). Substituting \((0, 0)\), we get \(1 > 0\), which is true.
• For \(x + 2y - 5 = 0\), the inequality is \(x + 2y - 5 < 0\). Substituting \((0, 0)\), we get \(-5 < 0\), which is also true.

Thus, the region \( R \) is given by:

• \(y > 3x + 1\)
• \(y < \frac{5-x}{2}\)

Next, substitute \( (a^2, a+1) \) into these inequalities to find \( a \).

• From \(y > 3x + 1\), substituting \(x = a^2\) and \(y = a + 1\), we get: \(a + 1 > 3a^2 + 1\). Simplifying gives:
• \(a + 1 - 3a^2 - 1 > 0 \ \Rightarrow \ -3a^2 + a > 0 \ \Rightarrow a(1 - 3a) > 0\)

Determine when \(a(1 - 3a) > 0\):

• This inequality holds when \(a\) is in the intervals \((- \infty, 0)\) or \(\left( \frac{1}{3}, \infty \right)\).

Next, consider the second inequality \( y < \frac{5-x}{2} \):

• Substituting \(x = a^2\) and \(y = a + 1\), we get: \(a + 1 < \frac{5 - a^2}{2}\)
• Simplifying gives: \(2(a + 1) < 5 - a^2 \ \Rightarrow \ 2a + 2 < 5 - a^2 \ \Rightarrow a^2 + 2a - 3 < 0\)

Factorize and solve the quadratic inequality:

• Solving \(a^2 + 2a - 3 = 0\) yields roots \(a = 1\) and \(a = -3\).
• The quadratic \(a^2 + 2a - 3\) is negative between root intervals \(-3 < a < 1\).

The solution to both inequalities is the intersection of \((- \infty, 0) \cup \left( \frac{1}{3}, \infty \right)\) and \((-3, 1)\):

• The intersection is \((-3, 0) \cup \left(-\frac{1}{3}, 1\right)\).

Thus, the correct set of values for \( a \) is:

Answer: \( (-3, 0) \cup \left(-\frac{1}{3}, 1\right) \)

Answer 2

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 : \( (-3, 0) \cup \left(\frac{1}{3}, 1\right) \)