Question

A function $f:A\to B$, where A = $\left\{x:-1\le x\le 1\right\}$ and B = $\left\{y:1\le y\le 2\right\}$ is defined by the rule $y=f\left(x\right)=1+x^2$.

Which of the following statement is correct?

• A. f is injective but not surjective
• B. f is surjective but not injective
• C. f is both injective and surjective
• D. f is neither injective nor surjective

Answer 1

The correct option is D

f is neither injective nor surjective

The explanation for the correct answer:

Step 1: Check the functionality one-one:

Given data,

A function $f:A\to B,f\left(x\right)=1+x^2$

A function is one-one for $x_{1,}{x}_2\in R,$ if

$f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow x_1=x_2$

Now, $1+{x_1}^2=1+{x^2}_2$

$$\begin{gathered} \Rightarrow {x_1}^2={x_2}^2 \\ \Rightarrow x_1=\sqrt{{x_2}^2} \\ \Rightarrow x_1=\pm x_2 \end{gathered}$$

So, the function is not one-one.

Let us consider, $f\left(1\right)=f\left(-1\right)=2$

It shows, $f$ is not a one-one function.

Step 2: Check the functionality onto:

Let us consider an element $-2$ in co-domain $B$. Then

$f\left(x\right)=1+x^2$

$\Rightarrow f\left(-2\right)=1+4$

$\Rightarrow f\left(-2\right)=5$

This implies that $f\left(x\right)=1+x^2$ is positive for all $x\in R$.

Thus, the element $-2$ in the co-domain $B$ is not image of any of the domain $A$.

Therefore, $f$ is not onto function.

Hence, option (D) is the correct answer.