Let $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The functions $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ are defined as $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$.
$(g \circ f)(1) = g(f(1)) = g(2) = 3$ [since $f(1) = 2$ and $g(2) = 3$]
$(g \circ f)(3) = g(f(3)) = g(5) = 1$ [since $f(3) = 5$ and $g(5) = 1$]
$(g \circ f)(4) = g(f(4)) = g(1) = 3$ [since $f(4) = 1$ and $g(1) = 3$]
Therefore,$g \circ f = \{(1,3), (3,1), (4,3)\}$.

Explore More

Similar Questions

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is:

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

Let $f$ be a composite function of $x$ defined by $f(u) = \frac{1}{u^2 + u - 2}$ and $u(x) = \frac{1}{x - 1}$. Then the number of points $x$ where $f$ is discontinuous is

Let $f(x) = 1 - x$, $g(x) = \frac{1}{1 - x}$, and $h(x) = \frac{1}{x}$ be three functions, for $x \neq 0, 1$. If a function $F(x)$ satisfies $f(F(h(x))) = g(x)$, then which of the following is true?

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f(g(x))=x$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo