**Educational Content on Functions**---### Diagram AnalysisThe image shows two diagrams that define functions \( H \) and \( K \) using arrow notations. Here's a detailed explanation:1. **Function \( H \):** - **Domain of \( H \)**: Set \( X = \{a, b, c\} \) - **Co-domain of \( H \)**: Set \( Y = \{d, e, f, g\} \) - **Mapping**: - \( a \rightarrow d \) - \( b \rightarrow e \) - \( c \rightarrow f \)2. **Function \( K \):** - **Domain of \( K \)**: Set \( X = \{a, b, c\} \) - **Co-domain of \( K \)**: Set \( Y = \{d, e, f, g\} \) - **Mapping**: - \( a \rightarrow d \) - \( b \rightarrow d \) - \( c \rightarrow f \)### Questions and Analysis**(a) Is \( H \) one-to-one? Why or why not?**- **Options:** 1. \( H \) is one-to-one because every element of \( X \) is sent by \( H \) to a different element of \( Y \): \( a \) is sent to \( d \), \( b \) is sent to \( f \), and \( c \) is sent to \( f \). 2. \( H \) is one-to-one because \( H(b) = f = H(c) \) and \( b \neq c \). 3. \( H \) is not one-to-one because \( g \in Y \) but \( g \not \in H(x) \) for any \( x \in X \). 4. \( H \) is not one-to-one because \( H(b) = f = H(c) \) and \( b \neq c \). 5. \( H \) is not one-to-one because \( e \in Y \) but \( e \not \in H(x) \) for any \( x \in X \).**Is \( H \) onto? Why or why not?**- **Options:** ### Educational Content on Function Properties#### Problem Set:**(a) Is \( H \) one-to-one? Why or why not?**- \( H \) is one-to-one because every element of \( X \) is sent by \( H \) to an element of \( Y \): \( a \) is sent to \( d \), \( b \) is sent to \( f \), and \( c \) is sent to \( f \).- \( H \) is one-to-one because \( H(b) = f = H(c) \) and \( b \neq c \).- \( H \) is not one-to-one because \( g \in Y \) but \( g \neq H(x) \) for any \( x \) in \( X \).- \( H \) is not one-to-one because \( H(b) = f = H(c) \).- \( H \) is not one-to-one because \( e \in Y \) but \( e \neq H(x) \) for any \( x \) in \( X \).**Is \( H \) onto? Why or why not?**- \( H \) is onto because \( H(a) = d \), \( H(b) = f \), and \( H(c) = f \).- \( H \) is onto because \( H(b) = f = H(c) \) and \( b + c \).- \( H \) is not onto because \( H(b) = f = H(c) \).- \( H \) is not onto because \( e \in Y \) but \( e \neq H(x) \) for any \( x \) in \( X \).- \( H \) is not onto because \( H(b) = f = H(c) \) and \( b + c \).**(b) Is \( K \) one-to-one? Why or why not?**- \( K \) is one-to-one because every element of \( X \) is sent by \( K \) to an element of \( Y \): \( a \) is sent to \( f \), \( b \) is sent to \( d \), and \( c \) is sent to \( e \).- \( K \) is one-to-one because \( g \in Y \) but \( g \neq K(x) \) for

Given: X=a, b, c and Y=d, e, f, g. The functions H and K are defined by arrow diagrams: We have to choose the correct answer for the given questions: Is H one-to-one? Why or why not?Is H onto? Why or why not? Is K one-to-one? Why or why not?Is K onto? Why or why not?We know that a function f : X → Y is said to be one-to-one if, for every y∈Y, there is at most one x∈X such that fx=y. And f is onto if, for every y∈Y, there exists at least one x∈X such that fx=y. (a) H is not one-to-one g∈Y but g≠Hx for any x in X.H is not one-to-one e∈Y but e≠Hx for any x in X. H is not onto e∈Y but e≠Hx for any x in X.(b) K is not one-to-one g∈Y but g≠Kx for any x in X. K is not onto g∈Y but g≠Kx for any x in X.