Question

The point $(a^2,a+1)$ is a point in the angle between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin, then?

• A. $a\ge 1$ or $a\le -3$
• B. $a\in (-3,0)\cup (3,1)$
• C. $a\in (0,1)$
• D. $a\in (-\infty ,0)$

Answer 1

The correct option is A $a\ge 1$ or $a\le -3$

We have,

Lines

$3x-y+1=0......\left(1\right)$

$x+2y-5=0......\left(2\right)$

And the points $(a^2,a+1)$ lies the line,

Then,

Equation (1):-

$3x-y+1=0$

$\Rightarrow 3a^2-(a+1)+1=0$

$\Rightarrow 3a^2-a-1+1=0$

$\Rightarrow 3a^2-a=0$

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

$\Rightarrow a=0,a=\frac{1}{3}$

Equation (2):-

$x+2y-5=0$

$\Rightarrow a^2+2(a+1)-5=0$

$\Rightarrow a^2+2a+2-5=0$

$\Rightarrow a^2+2a-3=0$

$\Rightarrow a^2+3a-a-3=0$

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

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

Now,

$a\ge 1anda\le -3$

Hence, this is the answer.