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 (a², a + 1) lies in R is

The correct option is (−3, 0) ∪ (1/3, 1).

1. Analysis of the region from the image:

As shown in the provided diagram, the two lines are:

$$L_1:x+2y-5=0$$

and

$$L_2:3x-y+1=0$$

The shaded region, denoted as
$R$
, contains the origin
$(0,0)$
. A point
$P(a^2,a+1)$
lies inside the region
$R$
if and only if it lies on the same side of both lines as the origin.

2. Position relative to line
$L_1$
:

Substituting the origin
$(0,0)$
into the expression for
$L_1$
:

$$0+2\left(0\right)-5=-5<0$$

Since the origin yields a negative value, the point
$(a^2,a+1)$
must also satisfy:

$$a^2+2(a+1)-5<0$$

Simplifying the inequality:

$$a^2+2a-3<0$$

Factoring the quadratic equation:

$$(a+3)(a-1)<0$$

This gives the first interval:

$$a\in (-3,1)$$

Let this be Inequality (1).

3. Position relative to line
$L_2$
:

Substituting the origin
$(0,0)$
into the expression for
$L_2$
:

$$3\left(0\right)-0+1=1>0$$

Since the origin yields a positive value, the point
$(a^2,a+1)$
must also satisfy:

$$3a^2-(a+1)+1>0$$

Simplifying the inequality:

$$3a^2-a>0$$

Factoring the quadratic expression:

$$a(3a-1)>0$$

This yields the second interval:

$$a\in (-\infty ,0)\cup (\frac{1}{3},\infty )$$

Let this be Inequality (2).

4. Finding the intersection:

To find the set of all values of
$a$
for which the point lies in the region
$R$
, we find the intersection of Inequality (1) and Inequality (2):

$$a\in (-3,1)\cap \left((-\infty ,0)\cup (\frac{1}{3},\infty )\right)$$

Taking the intersection segment by segment:
- The intersection of
$(-3,1)$
and
$(-\infty ,0)$
is
$(-3,0)$
.
- The intersection of
$(-3,1)$
and
$(\frac{1}{3},\infty )$
is
$(\frac{1}{3},1)$
.

Combining these sets yields the final solution:

$$a\in (-3,0)\cup (\frac{1}{3},1)$$