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:
- \( (-3, -1) \cup \left(-\frac{1}{3}, 1\right) \)
- \( (-3, 0) \cup \left(-\frac{1}{3}, 1\right) \)
- \( (-3, 0) \cup \left(\frac{2}{3}, 1\right) \)
- \( (-3, -1) \cup \left(\frac{1}{3}, 1\right) \)
The Correct Option is B
Approach Solution - 1
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: \( (-3, 0) \cup \left(-\frac{1}{3}, 1\right) \)
Approach Solution -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 : \( (-3, 0) \cup \left(\frac{1}{3}, 1\right) \)
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