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 lie in R, is: [2024]

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: [2024]

Smart Achievers · Accepted Answer

(4)

It is given that, region R lies between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0.

The point $(a^{2}, a + 1)$ and (0, 0) lie in the region R.

$\Rightarrow (a^{2}, a + 1)$ and (0, 0) are on same side of both the line.

For Line 3x – y + 1 = 0, O(0, 0) is on the right side of the line.

So, point $(a^{2}, a + 1)$ will also be on right side of the line.

$\Rightarrow 3 a^{2} - a - 1 + 1 > 0$

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

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

For line x + 2y – 5 = 0, O(0, 0) is on the left side of the line.

So, point $(a^{2}, a + 1)$ will also be on left side of the line.

$\Rightarrow a^{2} + 2 a + 2 - 5 < 0$

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

$\Rightarrow a \in (- 3, 1)$ ... (ii)

From the intersection of (i) and (ii), we get

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

Solution

Q.
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: [2024]

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

2 $(- 3, 0) \cup (\frac{2}{3}, 1)$

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

4 $(- 3, 0) \cup (\frac{1}{3}, 1)$

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Solution

Q. 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) lie in R, is: [2024]

1 (-3,-1)∪(13,1)

2 (-3,0)∪(23,1)

3 (-3,-1)∪(-13,1)