Is $g=\{(1,1),(2,3),(3,5),(4,7)\}$ a function? If $g$ is described by $g(x)=\alpha x+\beta$, then what value should be assigned to $\alpha$ and $\beta$ ?
Given that, $g=\{(1,1),(2,3),(3,5),(4,7)\}$.
Here, each element of domain has unique image. So, $g$ is a function.
Now given that,
$$\begin{aligned} g(x) & =\alpha x+\beta \\ g(1) & =\alpha+\beta \\ \alpha+\beta & =1 \quad\text{.... (i)}\\ g(2) & =2 \alpha+\beta \\ 2 \alpha+\beta & =3\quad\text{.... (ii)} \end{aligned}$$
From Eqs. (i) and (ii),
$$\begin{aligned} 2(1-\beta)+\beta & =3 \\ \Rightarrow\quad 2-2 \beta+\beta & =3 \\ \Rightarrow\quad 2-\beta & =3 \\ \beta & =-1 \\ \text{If}\quad \beta & =-1, \text { then } \alpha=2 \\ \alpha & =2, \beta=-1 \end{aligned}$$
Are the following set of ordered pairs functions? If so examine whether the mapping is injective or surjective.
(i) $\{(x, y): x$ is a person, $y$ is the mother of $x\}$.
(ii) $\{(a, b): a$ is a person, $b$ is an ancestor of $a\}$.
(i) Given set of ordered pair is $\{(x, y): x$ is a person, $y$ is the mother of $x\}$.
It represent a function. Here, the image of distinct elements of $x$ under $f$ are not distinct, so it is not a injective but it is a surjective.
(ii) Set of ordered pairs $=\{(a, b)$ : $a$ is a person, $b$ is an ancestor of $a\}$
Here, each element of domain does not have a unique image. So, it does not represent function.
If the mappings $f$ and $g$ are given by $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$, write $fog$.
$$\begin{aligned} \text{Given that,}\quad f & =\{(1,2),(3,5),(4,1)\} \\ \text{and}\quad g & =\{(2,3),(5,1),(1,3)\} \\ \text{Now,}\quad f \circ g(2) & =f\{g(2)\}=f(3)=5 \\ f \circ g(5) & =f\{g(5)\}=f(1)=2 \\ f \circ g(1) & =f\{g(1)\}=f(3)=5 \\ f \circ g & =\{(2,5),(5,2),(1,5)\} \end{aligned}$$
Let $C$ be the set of complex numbers. Prove that the mapping $f: C \rightarrow R$ given by $f(z)=|z|, \forall z \in C$, is neither one-one nor onto.
$$\begin{aligned} \text{The mapping}\quad & f: C \rightarrow R \\ \text{Given, }\quad & f(z)=|z|, \forall z \in C \\ & f(1)=|1|=1 \\ & f(-1)=|-1|=1 \\ & f(1)=f(-1) \\ \text{But }\quad & 1 \neq-1 \end{aligned}$$
So, $f(z)$ is not one-one. Also, $f(z)$ is not onto as there is no pre-image for any negative element of $R$ under the mapping $f(z)$.
Let the function $f: R \rightarrow R$ be defined by $f(x)=\cos x, \forall x \in R$. Show that $f$ is neither one-one nor onto.
Given function, $f(x)=\cos x, \forall x \in R$
Now, $$f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0$$
$$\begin{array}{lc} \Rightarrow & f\left(\frac{-\pi}{2}\right)=\cos \frac{\pi}{2}=0 \\ \Rightarrow & f\left(\frac{\pi}{2}\right)=f\left(\frac{-\pi}{2}\right) \\ \text { But } & \frac{\pi}{2} \neq \frac{-\pi}{2} \end{array}$$
So, $f(x)$ is not one-one.
Now, $f(x)=\cos x, \forall x \in R$ is not onto as there is no pre-image for any real number. Which does not belonging to the intervals $[-1,1]$, the range of $\cos x$.