CAT 2025 Slot 1 · Quant · Question 8

Both are parabolas, so each turning value comes straight from the vertex.

f opens upwards with vertex at x=2c,

so its minimum is f\left( 2c \right)=4{c}^{2}-8{c}^{2}+8c=-4{c}^{2}+8c.

g opens downwards with vertex at x=\frac{3c}{2},

so its maximum is g\left( \frac{3c}{2}\right)=-\frac{9{c}^{2}}{4}+\frac{9{c}^{2}}{2}-2c=\frac{9{c}^{2}}{4}-2c.

The condition is -4{c}^{2}+8c>\frac{9{c}^{2}}{4}-2c.

Multiplying by 4: -16{c}^{2}+32c>9{c}^{2}-8c, so 25{c}^{2}-40c<0 and 5c\left( 5c-8 \right)<0.

Hence 0<c<\frac{8}{5}. Of the options only \frac{1}{2} lies in this range.

Note that 2 just fails, since \frac{8}{5}=1.6.