Question

The point (a², a + 1) lies in the angle between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing the origin, if-

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

Answer 1

Answer: (A)

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

Answer 2

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

3a² – (a + 1) + 1 > 0

i.e. a(3a – 1) > 0

gives a Ī (– , 0) Č $\left( \frac { 1 } { 3 } , \infty \right)$

and a² + 2(a + 1) – 5 < 0

i.e. a² + 2a – 3 < 0

i.e. (a – 1) (a + 3) < 0

gives a Ī (–3, 1)

Intersection of the above inequalities gives

a Ī (–3, 0) Č $\left( \frac { 1 } { 3 } , 1 \right)$ .