Question

The least integral value of $a$ for which the graphs of the functions $y=2ax+1$ and $y=(a-6)x^2-2$ do not intersect is:

• A. -6
• B. -5
• C. 3
• D. 2

Answer 1

The correct option is B -5

For no intersection of graphs of functions, $y=2ax+1$ and $y=(a-6)x^2-2$, There should not any common points between two curves.

Putting the value of $y$ from equation of line into equation of given parabola, we get,

$\Rightarrow (2ax+1)=(a-6)x^2-2$

$\Rightarrow (a-6)x^2-\left(2a\right)x-3=0$ ...$\left(1\right)$

Equation $\left(1\right)$ is a quadratic equation in $x$.

For no intersection of both given functions, the equation $\left(1\right)$ must not have any real solutions.

A quadratic equation have no real roots if the value of it's discriminant is less than zero.

Hence $D=b^2-4ac<0$

$\Rightarrow D=\left(\right(-2a{)}^2)-4\times (a-6)\times (-3)<0$

$\Rightarrow D=4a^2+12a-72<0$

$\Rightarrow (a+6)(a-3)<0$

Hence Value of $a$ for the graphs of given functions do not intersect lies between $(-6,3)$

So the least integral value will be $(-5)$. Correct answer is $A$.