Question

Assertion : Let $A =\{-1,1,2,3\}$ and $B =\{1,4,9\}$, where $f : A \rightarrow B$ given by $f ( x )= x ^{2}$, then $f$ is a many-one function.
Reason : If $x_{1} \neq x_{2} \Rightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)$, 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 $f : R \rightarrow R$ 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.