CAT Exam 2021 Slot 1 Solved Paper

The quadratic equation of the form $\left|{\mathrm{x}}^2-4\mathrm{x}-13\right|=\mathit{r}$ has its minimum value at $\mathit{x}=-\frac{\mathit{b}}{2}\mathit{a}$ 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 : $\left|{\mathrm{x}}^2-4\mathrm{x}-13\right|$ as per the given condition, this must have 3 real roots.
The curve A B C D E represents the function $\left|{\mathrm{x}}^2-4\mathrm{x}-13\right|$ 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 $\mathit{x}=2$ $=2^2-8-13=-17$ the reflection about the x-axis is 17.
Alternatively
$\left|{\mathrm{x}}^2-4\mathrm{x}-13\right|=\mathrm{r}$
This can represented in two parts:
${\mathrm{x}}^2-4\mathrm{x}-13=\mathrm{r}$ if $\mathrm{r}$ is positive.
${\mathit{x}}^2-4\mathit{x}-13=-\mathit{r}$ if $\mathrm{r}$ is negative.
Considering the firs case : ${\mathrm{x}}^2-4\mathrm{x}-13=\mathrm{r}$
The quadratic equation becomes : ${\mathrm{x}}^2-4\mathrm{x}-13-\mathrm{r}=0$
The discriminant for this function is : ${\mathrm{b}}^2-4\mathrm{ac}=16-\left(4\left(-13-\mathrm{r}\right)\right)=68+4\mathrm{r}$
Since r is positive the discriminant is always greater than 0 this must have two distinct roots.
For the second case :
${\mathrm{x}}^2-4\mathrm{x}-13+\mathrm{r}=0$ the function inside the modulus is negative
The discriminant is $16-\left(4\left(\mathrm{r}-13\right)\right)=68-4\mathrm{r}$
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, $68-4\mathit{r}=0$
$\mathit{r}=17$ for $\mathit{r}=17$ we can have exactly 3 roots.