Question

Suppose $f\left(x\right)=ax+b$ and $g\left(x\right)=bx+a$, where $a$ and $b$ are positive integers. If $f\left(g\right(50\left)\right)-g\left(f\right(50\left)\right)=28$ then the product $\left(ab\right)$ can have the value equal to

• A. $12$
• B. $48$
• C. $180$
• D. $210$

Answer 1

Answer: (A), (D)

$12$
$210$

$f\left(g\right(x\left)\right)=a(bx+a)+b$
$=abx+a^2+b$
$g\left(f\right(x\left)\right)=b(ax+b)+a$
$=abx+b^2+a.$
Therefore $f\left(g\right(x\left)\right)-g\left(f\right(x\left)\right)$
$=abx+a^2+b-(abx+b^2+a)$
$=a^2-b^2+b-a$
$=(a-b)(a+b-1)$
Since it is a constant function.
$(a-b)(a+b-1)=28$
$(a-b)(a+b-1)=4\left(7\right)$
Hence $a-b=4$
$a+b=8$
$a=6$ and $b=2$
Hence $ab=12$
Also
$(a-b)(a+b-1)=\left(1\right)\left(28\right)$
$a-b=1$
$a+b=29$
$a=15$ and $b=14$
Therefore $ab=14\left(15\right)=210$