CAT 2021 Slot 2 · Quant · Question 13

Both quadratics are built from {x}^{2}+ 2x,

since 2{x}^{2}+ 4x + 9 = 2\left({x}^{2}+ 2x \right) + 9.

Put t ={x}^{2}+ 2x:

f = \frac{t+4}{2t+9}

Since a square is never negative,

t ={x}^{2}+ 2x ={\left( x+1 \right)}^{2}- 1 \ge -1

and t takes every value from −1 upwards.

Force the numerator to match the denominator:

f = \frac{t+4}{2t+9}= \frac{\frac{1}{2}\left( 2t+8 \right)}{2t+9}= \frac{1}{2}\left( 1 - \frac{1}{2t+9}\right)

At t=-1 (that is x=-1) we get f=\frac{3}{7}, which is attained. As t\to \infty, f\to \frac{1}{2}, which is never reached.

So the range is \left[ \frac{3}{7},\frac{1}{2}\right).