Question

The maximum negative integral value of $b$ for which the point $\left(2 b + 3 , b^{2}\right)$ 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, $y_{2} < y_{1}\Rightarrow \frac{3 x_{1} - a \left(a - 2\right)}{4} < y_{1}$
Putting $x_{1}=2b+3,y_{1}=b^{2}$ , we get,
$\Rightarrow 3\left(2 b + 3\right)-a\left(a - 2\right) < 4b^{2}.$
$\Rightarrow a^{2}-2a+4b^{2}-6b-9>0$
Now $\forall a\in R$ , $D < 0$
$\Rightarrow 4-4\left(4 b^{2} - 6 b - 9\right) < 0$
$\Rightarrow 1-4b^{2}+6b+9 < 0\Rightarrow 4b^{2}-6b+10>0$
$\Rightarrow 2b^{2}-3b-5>0\Rightarrow \left(2 b - 5\right)\left(b + 1\right)>0$
$\Rightarrow b\in \left(- \infty , - 1\right)\cup\left(\frac{5}{2} , \infty\right)$
Hence, the maximum negative integer value of $b$ is $-2$