Let f: A rightarrow B and g: B rightarrow C b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $g \circ f: A \rightarrow C$ is one-one. By definition,if $g(f(x_1)) = g(f(x_2))$,then $x_1 = x_2$ for all $x_1, x_2 \in A$. Suppose $f

Vedclass · Accepted Answer

$f$ is one-one and $g$ need not be one-one

Vedclass · Answer

(B) Given that $g \circ f: A \rightarrow C$ is one-one. By definition,if $g(f(x_1)) = g(f(x_2))$,then $x_1 = x_2$ for all $x_1, x_2 \in A$. Suppose $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Since $g \circ f$ is one-one,this implies $x_1 = x_2$. Thus,$f$ must be one-one. However,$g$ does not necessarily have to be one-one. For example,if $A = \{1\}$,$B = \{2, 3\}$,$C = \{4\}$,$f(1) = 2$,$g(2) = 4$,$g(3) = 4$,then $g \circ f(1) = 4$ is one-one,but $g$ is not one-one. Therefore,$f$ is one-one and $g$ need not be one-one.

Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be any two functions and $g \circ f: A \rightarrow C$ is one-one,then

Similar Questions

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\}$ is equal to

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}(x) = $

If $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$,then the function $g(x)$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=x^{3}+5$,then $(fog)^{-1}(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module