A function f : R → R defined as f(x) = x^2 - 4x + 5 is (A) injective but not surjective. ​

Question

 function \(f: R \rightarrow R\) defined as \(f(x) = x^2 - 4x + 5\) is

(A) injective but not surjective.

(B) surjective but not injective.

(C) both injective and surjective.

(D) neither injective nor surjective.

Answer 1

Correct option is : (D) neither injective nor surjective.

Given \(f : R \rightarrow R \ \ F(x) = x^2 - 4x + 5\)

For one-one

We known if f(x1) = f(x2)

\(\Rightarrow x 1 = x2 \)

Where \(x1, x2 \in R\)

\(\therefore \ f(x1)= x1{^2} - 4x1 + 5 \)  …(1)

\(f(x2) = x2{^2} - 4 x2 + 5\)   …(2)

equating (1) and (2)

\(x1{^2}- 4 x1 + 5 = x2{^2}- 4x2 + 5\)

\(x1{^2} - x2{^2} - 4x1 + 4x2 = 0\)

\((x1 - x2)(x1 + x2 )- 4(x1- x2) = 0\)

\(\Rightarrow (x1 - x2) (x1+ x2 -4) = 0\)

\(\Rightarrow x1 + x2 - 4 =0\)

\(\Rightarrow x1 = 4 - x2\)

\(\therefore \ x1 = 4 - x2\)

\(\therefore x1 \neq x2\)

\(\therefore \ f(x) \) in not one-one.

y = x² – 4x + 5

y = (x – 2)² + 5 – 4

y = (x – 2)² +1 

As (x – 2)² > 0

\(\Rightarrow y -1 \geq 0\)

\(\Rightarrow y \geq 1\)

\(\therefore \)  Range \((f) \in [1, \infty]\)

And codomain \(\in\) R

\(\therefore \)  Range \(\neq\) Codomain

\(\therefore \)  f(x) is not onto