Question

If $f\left(x\right)=x^2-2x;f:[0,3]\to A$, then $f$ is

• A. not an injective function
• B. a surjective function if $A=[-1,3]$
• C. an injective function
• D. a surjective function if $A=[0,3]$

Answer 1

Answer: (A), (B)

a surjective function if $A=[-1,3]$

Given: $f\left(x\right)=x^2-2x$
$\Rightarrow f\left(x\right)=(x-1{)}^2-1$


Clearly vertex is at $(1,-1)$
Since vertex lies in the given interval $[0,3]$, so the function is not an injective function.
Also, $f\left(3\right)=3;f\left(0\right)=0$
In the given interval, the least value is possible only at vertex.
$f(x{)}_{min}=f(1)=-1f(x{)}_{max}=f\left(3\right)=3$
$\therefore$Range of $f\in [-1,3]$
For $A=[-1,3],f$ is a surjective function.