Question

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

• A. $a\in (-3,0)\cup \left(\frac{1}{3},1\right)$
• B. $a\in (-\infty ,3)\cup \left(\frac{1}{3},1\right)$
• C. $a\in \left(-3,\frac{1}{3}\right)$
• D. $a\in \left(\frac{1}{3},\infty \right)$

Answer 1

The correct option is A

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

Explanation for the correct option.

The lines $3x-y+1=0$ and $x+2y-5=0$ contains the origin.

So,

$$\begin{gathered} 3\times 0-0+1=1>0 \\ 0+2\times 0-5=-5<0 \end{gathered}$$

Also, the point $(a^2,a+1)$ lies in the angle between the lines $3x-y+1=0$ and $x+2y-5=0$.

So,

$$\begin{gathered} 3\times a^2-\left(a+1\right)+1>0 \\ \Rightarrow 3a^2-a>0 \\ \Rightarrow a\left(3a-1\right)>0 \\ \therefore a\in \left(-\infty ,0\right)\cup \left(\frac{1}{3},\infty \right)...\left(1\right) \\ {a}^2+2\times \left(a+1\right)-5<0 \\ \Rightarrow a^2+2a-3<0 \\ \Rightarrow a^2+3a-a-3<0 \\ \Rightarrow a\left(a+3\right)-\left(a+3\right)<0 \\ \Rightarrow \left(a+3\right)\left(a-1\right)<0 \\ \therefore a\in \left(-3,1\right)...\left(2\right) \end{gathered}$$

From (1) and (2),

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

Hence, the correct option is (A).