Match the following $ \begin{array}{|c|c|c|} \hline & \text{List I} & \text{List II} \\ \hline \text{(A)} &…
Match the following
$
\begin{array}{|c|c|c|}
\hline & \text{List I} & \text{List II} \\
\hline \text{(A)} & f: \mathbb{R} \rightarrow \mathbb{R} \text{ is such that } f(x)=p x+q, (p \neq 0), \forall x \in \mathbb{R} & I. f \text{ is neither one-one nor onto} \\
\hline \text{(B)} & f: \mathbb{R} \rightarrow \mathbb{R}^{+} \cup\{0\} \text{ is such that } f(x)=x^{2}, \forall x \in \mathbb{R} & II. f \text{ is both one-one and onto} \\
\hline \text{(C)} & f: \mathbb{N} \rightarrow \mathbb{N} \text{ is such that } f(n)=n^{2}+2 n+3, \forall n \in \mathbb{N} & III. f \text{ is one-one but not onto} \\
\hline \text{(D)} & f: \mathbb{R} \rightarrow \mathbb{R} \text{ is such that } f(x)=2\left(\cos^{2} 5 x+\sin^{2} 5 x\right), \forall x \in \mathbb{R} & IV. f \text{ is onto but not one-one} \\
\hline && V. f \text{ is a constant function and also a bijection} \\
\hline
\end{array}$
The correct answer is
$\begin{array}{cccc} A & B & C & D \\ II & IV & III & I \end{array}$
$\begin{array}{cccc} A & B & C & D \\ II & IV & V & I \end{array}$
$\begin{array}{cccc} A & B & C & D \\ II & I & III & V \end{array}$
$\begin{array}{cccc} A & B & C & D \\ III & II & I & IV \end{array}$
Solution
(A) For function \(f: R \rightarrow R\) is defined as \(f(x)=p x+q,(p \neq 0)\) is a linear function.] And linear functions are one-one and onto in set of real numbers \((R)\).
So, \(\mathrm{A} \rightarrow \mathrm{II}\)
(B) For function \(f: R \rightarrow R^{+} \cup\{0\}\) is defined as
\(f(x)=x^2\)
\(\because f(-1)=f(1)=1\), so \(f(x)\) is not one-one function but range of \(f(x)=x^2\) is \([0, \infty)\),
\(\because \quad x^2 \geq 0, \forall x \in R \text {. }\)
So, \(f\) is onto but not one-one.
So, \(\mathrm{B} \rightarrow\) IV
(C) For \(f: N \rightarrow N\) is defined as \(f(x)=n^2+2 n+3\) is one-one but not onto because there is not value of \(n\), for which \(f(n)=3\).
So, \(\mathrm{C} \rightarrow\) III
(D) For \(f: R \rightarrow R\) is defined as \(f(x)=2\)
\(\left(\cos ^2 5 x+\sin ^2 5 x\right)=2(1)=2\)
\(\because f\) is define for every value of \(x \in R\), but range of \(f\) is \(\{2\}\).
So, \(f\) is neither one-one nor onto.
Hence, option (a) is correct.