1. Identify the required region $R$

The two lines are: $3x-y+1=0$ $x+2y-5=0$

We are told to take the interior region between these lines containing the origin.

So first determine on which side of each line the origin lies.

• For $3x-y+1=0$, at $(0,0)$: $3(0)-0+1=1>0$ Hence the origin lies in the half-plane $3x-y+1>0$

• For $x+2y-5=0$, at $(0,0)$: $0+0-5=-5<0$ Hence the origin lies in the half-plane $x+2y-5<0$

Therefore, $R=\{(x,y): 3x-y+1>0 \text{ and } x+2y-5<0\}$

2. Substitute the point $(a^2,a+1)$ into these inequalities

We need $(x,y)=(a^2,a+1)$ to satisfy both inequalities.

First inequality

$3x-y+1>0$ Substitute $x=a^2,\ y=a+1$: $3a^2-(a+1)+1>0$ $3a^2-a>0$ $a(3a-1)>0$

This gives $a<0 \quad \text{or} \quad a>\frac13$

Second inequality

$x+2y-5<0$ Substitute $x=a^2,\ y=a+1$: $a^2+2(a+1)-5<0$ $a^2+2a-3<0$ $(a+3)(a-1)<0$

This gives $-3<a<1$

3. Take the intersection

We need both conditions simultaneously: $\left(a<0 \text{ or } a>\frac13\right) \quad \text{and} \quad -3<a<1$

Intersecting:

• From $a<0$ with $-3<a<1$, we get $(-3,0)$

• From $a>\frac13$ with $-3<a<1$, we get $\left(\frac13,1\right)$

Thus the required set is $(-3,0)\cup\left(\frac13,1\right)$

4. Match with options

This is Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.