CAT 2023

Slot 3

QA

Question & Solution

It is given that for some real numbers a and b, the system of equations $x + y = 4$ and $(a+5)x+(b^2-15)y=8b$ has infinitely many solutions for x and y.

Hence, we can say that 

=> $\ \frac{\ a+5}{1}=\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}$

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=> $\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}\ =>\ b^2-2b-15=0$

=> $\left(b-5\right)\left(b+3\right)=0$

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is $a+5=b^2-15\ =>\ a\ =b^2-20$

When $b=5,a=5^2-20=5$

When $b=-3,a=3^2-20=-11$

The maximum value of $ab=(-3)\cdot(-11)=33$

The correct option is A