CAT Exam 2021 Slot 2 Solved Paper

$\mathit{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+2\mathrm{x}+4}{2{\mathrm{x}}^2+4\mathrm{x}+9}$
If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the and ${\mathrm{x}}^2$ in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.
$\mathit{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+2\mathrm{x}+4}{2{\mathrm{x}}^2+4\mathrm{x}+9}$ = $\frac{{\mathrm{x}}^2+2\mathrm{x}+4.5-0.5}{2{\mathrm{x}}^2+4\mathrm{x}+9}$
$\frac{{\mathrm{x}}^2+2\mathrm{x}+4.5}{2{\mathrm{x}}^2+4\mathrm{x}+9}-\frac{0.5}{2{\mathrm{x}}^2+4\mathrm{x}+9}$
$=\frac{1}{2}-\frac{0.5}{2{\mathrm{x}}^2+4\mathrm{x}+9}$
Now, we only have terms of x in the denominator.
The maximum value of the expression is achieved when the quadratic expression $2{\mathrm{x}}^2+4\mathrm{x}+9$ achieves its highest value, that is infinity.
In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ')'.
To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.
The minimum value is obtained when $4\mathit{x}+4=0\left[\frac{\mathit{d}}{\mathit{d}}\mathit{x}=0\right]$
The expression comes as 7.
The entire expression becomes $\frac{3}{7}$
Hence,$\left[\frac{3}{7},\frac{1}{2}\right]$