CAT 2021 Quant Question [Slot 1] with Solution 22

Option: 1
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 :
\({x^2} - 4x - 13\; = \;r\;if\;r\;is\;positive.\)
\({x^2} - 4x - 13\; = \; - r\;if\;r\;is\;negative.\)
Considering the first case : \({x^2} - 4x - 13\; = r\)
The quadratic equation 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.

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