Match the following \(\begin{array}{|c|c|c|} \hline & \text { List I } & \text { List II } \ \hline \text { (A) } & \begin{array}{l} f: R \rightarrow R \text { is such that } f(x)=p x+q \ (p \neq 0), \forall x \in R \end{array} & \begin{array}{l} \text {I. } f \text { is neither one-one nor onto } \end{array} \ \hline \text { (B) } & \begin{array}{l} f: R \rightarrow R^{+} \cup\{0\} \text { is such that } f(x)=x^2, \forall x \in R \end{array} & \begin{array}{l} \text {II. } f \text { is both one-one and onto } \end{array} \ \hline \text { (C) } & \begin{array}{l} f: N \rightarrow N \text { is such that } f(n)=n^2+2 n+3, \forall n \in N \end{array} & \begin{array}{l} \text {III. } f \text { is one-one but not onto } \end{array} \ \hline \text { (D) } & \begin{array}{l} f: R \rightarrow R \text { is such that } f(x)=2\left(\cos ^2 5 x+\sin ^2 5 x\right) \ \forall x \in R \end{array} & \begin{array}{l} \text {IV. } f \text { is onto but not one-one } \end{array} \ \hline && V. f \text{ is a constant function and also a bijection} \ \hline \end{array}\) The correct answer is

Question

Match the following \(\begin{array}{|c|c|c|} \hline & \text { List I } & \text { List II } \ \hline \text { (A) } & \begin{array}{l} f: R \rightarrow R \text { is such that } f(x)=p x+q \ (p \neq 0), \forall x \in R \end{array} & \begin{array}{l} \text {I. } f \text { is neither one-one nor onto } \end{array} \ \hline \text { (B) } & \begin{array}{l} f: R \rightarrow R^{+} \cup\{0\} \text { is such that } f(x)=x^2, \forall x \in R \end{array} & \begin{array}{l} \text {II. } f \text { is both one-one and onto } \end{array} \ \hline \text { (C) } & \begin{array}{l} f: N \rightarrow N \text { is such that } f(n)=n^2+2 n+3, \forall n \in N \end{array} & \begin{array}{l} \text {III. } f \text { is one-one but not onto } \end{array} \ \hline \text { (D) } & \begin{array}{l} f: R \rightarrow R \text { is such that } f(x)=2\left(\cos ^2 5 x+\sin ^2 5 x\right) \ \forall x \in R \end{array} & \begin{array}{l} \text {IV. } f \text { is onto but not one-one } \end{array} \ \hline && V. f \text{ is a constant function and also a bijection} \ \hline \end{array}\) The correct answer is, $\begin{array}{cc} & A & B & C & D \ & II & IV & III & I \ \end{array}$, $\begin{array}{cc} & A & B & C & D \ & II & IV & V & I \ \end{array}$, $\begin{array}{cc} & A & B & C & D \ & II & I & III & V \ \end{array}$, $\begin{array}{cc} & A & B & C & D \ & III & II & I & IV \ \end{array}$

MARKS App · Accepted Answer

(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.

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