Suppose `f(x)=a x+ba n dg(x)=b x+a ,w h e r eaa n db` are positive integers. If `f(g(20))-g(f(20))=28 ,` then which of the following is not true? `a=15` b. `a=6` c. `b=14` d. `b=3`

Options

A. `a=15`
B. `a=6`
C. `b=14`
D. `b=3`

Answer

D

Explanation

`f(g(x))=a(bx+a)+b`
`=abx+a^(2)+b" (i)"`
`g(f(x))=b(ax+b)+a`
`=abx+b^(2)+a" (ii)"`
From (i) - (ii), we get
`f(g(20))-g(f(20))=a^(2)-b^(2)+b-a`
`therefore" "(a^(2)-b^(2))+(b-a)=28`
`therefore" "(a-b)(a+b-1)=28=1xx 2x or 2 xx 14 or 4xx7`
If `a-b=1 and a+b-1=28`
Then `a=15, b=14`
If `a-b=2 and a+b-1=14` (not possible)
If `a-b=4 and a+b-1 =7`
Then a = 6 and b = 2