Question

Assertion : Let $A=\{-1,1,2,3\}$ and $B=\{1,4,9\}$, where $f:A\to B$ given by $f(x)=x^2$, then $f$ is a many-one function.
Reason : If $x_1\ne x_2\Rightarrow f\left(x_1\right)\ne f\left(x_2\right)$, for every $x_1,x_2\in$ domain, then $f$ is one-one or else many-one.

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct.

Answer 1

Answer: (D)

Function $f:R\to R$ is defined as $f(x)=x^4$
Let $x,y\in R$ such that $f(x)=f(y)$
$\Rightarrow x^4=y^4$
$\Rightarrow x=\pm y($ considering only real values $)$
Therefore, $f\left(x_1\right)=f\left(x_2\right)$ does not imply that $x_1=x_2$
For instance, $f(1)=f(-1)=1$
Therefore, $f$ is not one-one.
Consider an element $-2$ in codomain $R$. It is clear that there does not exist any $x$ in domain $R$ such that $f(x)=-2$
Therefore, $f$ is not onto.
Hence, function $f$ is neither one-one nor onto.