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Question

Question: Find the value of \[{f^{ - 1}}\left( x \right)\] if let \[f:R \to R\] be defined by \[f\left( x \rig...

Find the value of f1(x){f^{ - 1}}\left( x \right) if let f:RRf:R \to R be defined by f(x)=2x+6f\left( x \right) = 2x + 6 which is bijective mapping.
A) x23\dfrac{x}{2} - 3
B) 2x+62x + 6
C) x3x - 3
D) 6x+26x + 2

Explanation

Solution

Hint: we are given that f:RRf:R \to R is bijective mapping. First find x in terms of y by f(x)f\left( x \right)= y. After this you will get f1(y){f^{ - 1}}\left( y \right), but we are asked to find f1(x){f^{ - 1}}\left( x \right) which is found by simply replacing y by x in f1(y){f^{ - 1}}\left( y \right).

Complete step-by-step answer:
Let y=f(x)y = f\left( x \right)…………..(1)
\Rightarrow2x = y – 6
\Rightarrow $$$$x = \dfrac{y}{2} - 3= f1(y){f^{ - 1}}\left( y \right) (\because equation (1) can be written in form of x= f1(y){f^{ - 1}}\left( y \right))
Now replace y with x to find out f1(x){f^{ - 1}}\left( x \right)
Hence, f1(x){f^{ - 1}}\left( x \right)=x23\dfrac{x}{2} - 3.
∴ The correct option is ‘A’.

Note: Bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set.