Question

Suppose $f\left(x\right)=ax+b$ and $g\left(x\right)=bx+a,$ where $a,b$ are positive integers $a>b$. If $f\left(g\right(50\left)\right)-g\left(f\right(50\left)\right)=28,$ then the possible values of $ab$ is/are

• A. $210$
• B. $12$
• C. $240$
• D. $280$

Answer 1

Answer: (A), (B)

$12$

Given : $f\left(x\right)=ax+b$ and $g\left(x\right)=bx+a$
$f\left(g\right(x\left)\right)=a(bx+a)+b\Rightarrow f\left(g\right(x\left)\right)=abx+a^2+bg\left(f\right(x\left)\right)=b(ax+b)+a\Rightarrow g\left(f\right(x\left)\right)=abx+b^2+a\Rightarrow f\left(g\right(x\left)\right)-g\left(f\right(x\left)\right)=abx+a^2+b-(abx+b^2+a)f\left(g\right(50\left)\right)-g\left(f\right(50\left)\right)=a^2-b^2+b-a\Rightarrow (a-b)(a+b-1)=28$
$\therefore (a-b)(a+b-1)=1\times 28\text{or}2\times 14\text{or}4\times 7$

$\text{(i) Let}a-b=1,a+b-1=28\therefore a=15,b=14=ab=210$
If we consider the other case by interchanging the values, i.e
$a-b=28,a+b-1=1$ we wont be getting $a,b$ as positive integers.

$\text{(ii) Let}a-b=2,a+b-1=14$
Not possible as $a,b$ are positive integers

$\text{(iii) Let}a-b=4,a+b-1=7\therefore a=6,b=2\Rightarrow ab=12$
If we consider the other case by interchanging the values, i.e
$a-b=7,a+b-1=4$ we wont be getting $a,b$ as positive integers.