Question

The maximum negative integral value of $b$ for which the point $\left(2b+3,b^2\right)$ lies above the line $3x-4y-a\left(a-2\right)=0,\forall a\in R$ is

• A. $-1$
• B. $-3$
• C. $-2$
• D. $-4$

Answer 1

Answer: (C)

Let $\left(x_1,y_2\right)$ lie on the line
$\therefore 3x_1-4y_2-a\left(a-2\right)=0\Rightarrow 4y_2=3x_1-a\left(a-2\right)$
Now, $y_2<y_1\Rightarrow \frac{3x_1-a\left(a-2\right)}{4}<y_1$
Putting $x_1=2b+3,y_1=b^2$ , we get,
$\Rightarrow 3\left(2b+3\right)-a\left(a-2\right)<4b^2.$
$\Rightarrow a^2-2a+4b^2-6b-9>0$
Now $\forall a\in R$ , $D<0$
$\Rightarrow 4-4\left(4b^2-6b-9\right)<0$
$\Rightarrow 1-4b^2+6b+9<0\Rightarrow 4b^2-6b+10>0$
$\Rightarrow 2b^2-3b-5>0\Rightarrow \left(2b-5\right)\left(b+1\right)>0$
$\Rightarrow b\in \left(-\infty ,-1\right)\cup \left(\frac{5}{2},\infty \right)$
Hence, the maximum negative integer value of $b$ is $-2$