Question

The point (a², a + 1) lies in the angle between the line

3x – y + 1= 0 and x + 2y – 5 = 0 containing the origin if –

• A. aĪ (– 3, 0) Ē $\left( \frac{1}{3},1 \right)$
• B. a Ī (– , –3) Č$\left( \frac{1}{3},1 \right)$
• C. $a \in \left( - 3,\frac{1}{3} \right)$
• D. $a \in \left( \frac{1}{3},\infty \right)$

Answer 1

Answer: (A)

aĪ (– 3, 0) Ē $\left( \frac{1}{3},1 \right)$

Answer 2

Since origin and point (a², a + 1) lie on the same side of both the lines , so

3a² – (a + 1) + 1 > 0, a (3a – 1) > 0 gives

a ฮ (– ฅ, 0) ศ $\left( \frac{1}{3},\infty \right)$

and a² + 2(a + 1) – 5 < 0

a² + 2a – 3 < 0  (a – 1) (a + 3) < 0  a ฮ (–3, 1)

By both the inequalities a ฮ (–3, 0) ศ$\left( \frac{1}{3},1 \right)$