If the function f (x) = ax + b has its own inverse then the ordered pair (a, b) can be

Question

If the function f (x) = ax + b has its own inverse then the ordered pair (a, b) can be (A) (1, 0) (B) (-1, 0) (C) (-1, 1) (D) (1, 1)

Tardigrade Admin · Accepted Answer

y =f( x ) ⇒ x =f-1( x ) text now y = ax + b x =( y / a )-( b / a ) f-1( y )=( y / a )-( b / a ) f-1( x )=( x / a )-( b / a )....(1) text and f ( x )= ax + b ldots .(2) text now in order that (1) and (2) coincide a =(1/ a ) ....(1) ( b / a )=- b....(2) text from (1), a 2=1 ⇒ a =1 text or -1 text if a =-1, b = b ⇒ b ∈ R text if a=+1 text , then 2 b=0 ⇒ b=0 text hence (-1, R ),(1,0)

Q. If the function f(x)=ax+b has its own inverse then the ordered pair(a,b) can be

(1, 0)

(-1, 0)

(-1, 1)

(1, 1)

y=f(x)⇒x=f−1(x) now y=ax+b x=ay​−ab​ f−1(y)=ay​−ab​ f−1(x)=ax​−ab​....(1) and f(x)=ax+b….(2) now in order that (1) and (2) coincide a=a1​....(1) ab​=−b....(2) from (1), a2=1⇒a=1 or −1 if a=−1,b=b⇒b∈R if a=+1, then 2b=0⇒b=0 hence (−1,R),(1,0)

Q. If the function $f (x) = a x + b$ has its own inverse then the ordered pair $(a, b)$ can be