JEE Mathematics 2024 Question with Solution

Given: The lines are $3x - y + 1 = 0$ and $x + 2y - 5 = 0$. The point is $(a^2, a + 1)$.

Find: Values of $a$ for which the point lies in the region $R$ containing the origin.

First identify the side of each line that contains the origin.

For $3x - y + 1 = 0$, substituting $(0,0)$ gives $1 > 0$, so the required half-plane is

that is,

For $x + 2y - 5 = 0$, substituting $(0,0)$ gives $-5 < 0$, so the required half-plane is

that is,

Now substitute $x = a^2$ and $y = a + 1$.

From $3x - y + 1 > 0$,

Hence,

From $x + 2y - 5 < 0$,

Hence,

Taking the intersection,

Therefore, the working in the solution gives the interval $(-3,0) \cup \left(\frac{1}{3},1\right)$, while the listed option B is written as $(-3,0) \cup \left(-\frac{1}{3},1\right)$. The solution itself is inconsistent, but it explicitly marks B as the correct option. Hence the defensible answer from the source is B.