Question

If the points $(\alpha ,{\alpha }^2)$ and $(1,2)$ lie in the same angular region between the lines $3x+4y-1=0$ and $2x+y-3=0$, then the range of $\alpha$ is:

• A. $\alpha <-3or\alpha >1$
• B. $-3\le a\le 1$
• C. $\alpha <\frac{1}{2}or\alpha >1$
• D. none of these

Answer 1

The correct option is A $\alpha <-3or\alpha >1$

The Given Lines are $3x+4y-1=0$ and $2x+y-3=0$

Putting point $(1,2)$ in both lines we get,

$\Rightarrow 3\left(1\right)+4\left(2\right)-1=10>0$ and $2\left(1\right)+\left(2\right)-3=1>0$

Hence Point $(1,2)$ is same side of both the lines, opposite to origin.

If the point $(\alpha ,{\alpha }^2)$ also lies in same angular region as of $(1,2)$, then it must give positive values out of both the equations.

Hence, $3\alpha +4{\alpha }^2-1>0$ ...$\left(1\right)$

$2\alpha +{\alpha }^2-3>0$ ...$\left(2\right)$

From equation $\left(1\right)$ we get,

$\to (\alpha +1)(4\alpha -1)>0$

$\to (\alpha +3)(\alpha -1)>0$

On combining the result we get, $\alpha <-3or\alpha >1$

Hence correct option is $A$.