Question

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

	List I		List I
(A)	$f:R\to R$ defined by $f(x)=\cos (112x-37)$	(I)	Injection but not surjection
(B)	$f:A\to B$ defined by $f(x)=x\mid x$ when $A=[-2,2]$ and $B=[-4,4]$	(II)	Surjection but not injection
(C)	$f:R\to R$ defined by $f(x)=(x-2)(x-3)(x-5)$	(III)	Bijection
(D)	$f:N\to N$ 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)

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