Match the following: (A) f: R rightarrow | vedclass

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(A) For the function $f: R ightarrow R$ defined by $f(x)=px+q$ $(p 
eq 0)$,it is a linear function. Linear functions are bijections (both one-one a

Vedclass · Accepted Answer

$(A)-II, (B)-IV, (C)-III, (D)-I$

Vedclass · Answer

(A) For the function $f: R ightarrow R$ defined by $f(x)=px+q$ $(p 
eq 0)$,it is a linear function. Linear functions are bijections (both one-one and onto) on the set of real numbers $R$. Thus,$A ightarrow II$.$(B)$ For the function $f: R ightarrow R^{+} \cup \{0\}$ defined by $f(x)=x^2$,we observe that $f(-1)=f(1)=1$,so it is not one-one. However,for every $y \in R^{+} \cup \{0\}$,there exists $x = \sqrt{y} \in R$ such that $f(x)=y$,so it is onto. Thus,$B ightarrow IV$.$(C)$ For $f: N ightarrow N$ defined by $f(n)=n^2+2n+3$,it is one-one because $f(n_1)=f(n_2) \implies n_1^2+2n_1+3 = n_2^2+2n_2+3 \implies (n_1-n_2)(n_1+n_2+2)=0$,which implies $n_1=n_2$ for $n \in N$. It is not onto because for $f(n)=3$,$n^2+2n+3=3 \implies n(n+2)=0$,which gives $n=0$ or $n=-2$,neither of which is in $N$. Thus,$C ightarrow III$.$(D)$ For $f: R ightarrow R$ defined by $f(x)=2(\cos^2 5x + \sin^2 5x) = 2(1) = 2$. This is a constant function. $A$ constant function is neither one-one nor onto (unless the domain and codomain are singletons). Thus,$D ightarrow I$.Therefore,the correct matching is $(A)-II, (B)-IV, (C)-III, (D)-I$.

$(A)$ $f: R \rightarrow R$ is such that $f(x)=px+q$ $(p \neq 0)$,$\forall x \in R$	$I.$ $f$ is neither one-one nor onto
$(B)$ $f: R \rightarrow R^{+} \cup\{0\}$ is such that $f(x)=x^2$,$\forall x \in R$	$II.$ $f$ is both one-one and onto
$(C)$ $f: N \rightarrow N$ is such that $f(n)=n^2+2n+3$,$\forall n \in N$	$III.$ $f$ is one-one but not onto
$(D)$ $f: R \rightarrow R$ is such that $f(x)=2(\cos ^2 5x+\sin ^2 5x)$ $\forall x \in R$	$IV.$ $f$ is onto but not one-one
	$V.$ $f$ is a constant function and also a bijection

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