CAT 2021

Slot 1

QA

Question & Solution

The quadratic equation of the form $\mid x^2 - 4x - 13 \mid = r$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $\left|x^2-4\cdot x-13\right|$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $\left|x^2-4\cdot x-13\right|$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC ' DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $2^2-8-13$

= -17, the reflection about the x-axis is 17.

Alternatively,

$\mid x^2 - 4x - 13 \mid = r$ .

This can represented in two parts :

$x^2-4x-13\ =\ r \textit{ (if r is positive)}$

$x^2-4x-13\ =\ -r\textit{ (if r is negative)}$

Considering the first case : $x^2-4x-13\ =r$

The quadraticequation becomes : $x^2-4x-13-r\ =\ 0$

The discriminant for this function is : $b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$x^2-4x-13+r\ =\ 0$ the function inside the modulus is negaitve

The discriminant is $16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $\ 68-4r\ =\ 0$

r = 17, for r = 17 we can have exactly 3 roots.