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

Answer: (B)

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

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