Question 11 Mark
Assertion (A): The modulus function $f : R \rightarrow R$ given by $f ( x )=| x |$ is neither one-one nor onto.
Reason (R): The signum function $f : R \rightarrow R$ given by $f ( x )=\left\{\begin{array}{cl}1, & x>0 \\ 0, & x=0 \\ -1, & x<0\end{array}\right.$ is bijective.
Reason (R): The signum function $f : R \rightarrow R$ given by $f ( x )=\left\{\begin{array}{cl}1, & x>0 \\ 0, & x=0 \\ -1, & x<0\end{array}\right.$ is bijective.
Answer
View full question & answer→(c) A is true but R is false.
Explanation: Assertion: Here, $f : R \rightarrow R$ is given by
$f ( x )=| x |=\left\{\begin{array}{cc}x, & \text { if } x \geq 0 \\ -x, & \text { if } x<0\end{array}\right.$
It is seen that $f(-1)=|-1|=1, f(1)=|1|=1$
Therefore, $f(-1)=f(1)$ but $-1 \neq 1$
Therefore, f is not one-one.
Now, consider -1 ∈ R
It is known that f(x) = |x| is always non-negative
Thus, there does not exist any element x in domain R such that f (x) = |x| = -1.
Therefore, f is not onto.
Hence, the modulus function is neither one-one nor onto.
Reason: $f : R \rightarrow R , f ( x )=\left\{\begin{array}{ccc}1, & \text { if } & x>0 \\ 0, & \text { if } & x=0 \\ -1, & \text { if } & x<0\end{array}\right.$
It is seen that $f(1)=f(2)=1$ but $1 \neq 2$.
Therefore, f is not one-one
Now, as f(x) takes only three values (1, 0 or -1), therefore for the element -2 in codomain R, there does not exist any x in domain R such that f(x) = -2
Therefore, f is not onto. Hence, the Signum function is neither one-one nor onto.
Explanation: Assertion: Here, $f : R \rightarrow R$ is given by
$f ( x )=| x |=\left\{\begin{array}{cc}x, & \text { if } x \geq 0 \\ -x, & \text { if } x<0\end{array}\right.$
It is seen that $f(-1)=|-1|=1, f(1)=|1|=1$
Therefore, $f(-1)=f(1)$ but $-1 \neq 1$
Therefore, f is not one-one.
Now, consider -1 ∈ R
It is known that f(x) = |x| is always non-negative
Thus, there does not exist any element x in domain R such that f (x) = |x| = -1.
Therefore, f is not onto.
Hence, the modulus function is neither one-one nor onto.
Reason: $f : R \rightarrow R , f ( x )=\left\{\begin{array}{ccc}1, & \text { if } & x>0 \\ 0, & \text { if } & x=0 \\ -1, & \text { if } & x<0\end{array}\right.$
It is seen that $f(1)=f(2)=1$ but $1 \neq 2$.
Therefore, f is not one-one
Now, as f(x) takes only three values (1, 0 or -1), therefore for the element -2 in codomain R, there does not exist any x in domain R such that f(x) = -2
Therefore, f is not onto. Hence, the Signum function is neither one-one nor onto.
