Question

Match the functions of List-I with their nature in List-II and choose the correct option.

	List I		List I
(A)	$f: R \rightarrow R$ defined by $f(x)=\cos (112 \,x-37)$	(I)	Injection but not surjection
(B)	$f: A \rightarrow B$ defined by $f(x)=x \mid x$ when $A=[-2,2]$ and $B=[-4,4]$	(II)	Surjection but not injection
(C)	$f: R \rightarrow R$ defined by $f(x)=(x-2)(x-3)(x-5)$	(III)	Bijection
(D)	$f: N \rightarrow N$ defined by $f(n)=n+1$	(IV)	Neither injection nor surjection
		(V)	Composite function

Then, the correct match is

• A. A-I, B- II , C- III , D-IV
• B. A-IV, B- I , C- II , D-III
• C. A-IV, B- III , C- II , D-V
• D. A-IV, B- III , C- II , D-I

Answer 1

Answer: (D)

$(A) f: R \rightarrow R$
$f(x)=\cos (112 x-37)$
$f(x)$ is periodic function.
$\therefore $ It is not injective.
Range of $f(x)$ is $[-1,1]$
$\therefore $ It is also not surjective.
$\therefore A \rightarrow IV$
$ (B) f: A \rightarrow B, A \in[-2,2]$ and $ B \in[-4,4] $
$f(x)=x|x| $
$f(x)=\begin{cases}-x^{2} & -2 \leq x<0 \\ x^{2} & 0 \leq x \leq 2\end{cases}.$
Clearly $f(x)$ is bijective function.
$\therefore B \rightarrow III$
$(C) f: R \rightarrow R$
$f(x)=(x-2)(x-3)(x-5) $
$f(2)=f(3)=f(5)=0$
$\therefore $ It is not injective.
Range of $f(x)=R$
$\therefore $ It is surjective.
$\therefore C \rightarrow II$
$(D) f: N \rightarrow N$
$f(x)=n+1$
Clearly it is injective.
Range of $f(x)=\{2,3,4, \ldots\}$
Range $\neq$ Codomain
$\therefore $ It is not surjective.
$\therefore D \rightarrow I$