Column-1 gives functions and column-2 the nature of the functions.Column-1Column-2A) f:(0,∞)→(0,∞);f(x)=x1+xp) One-one, on toB) f:R−{0}→R,f(x)=x−1xq)One-one but not on toC) f:R−{0}→R,f(x)=x+1xr)On to but not one-oneD) f:R→R,f(x)=2x+sinxs)Neither one-one nor on to

A) f(x)=x1+x⇒f1(x)=1(1+x)2>0 B) f1(x)=1+1x2 C) f1(x)=2+cos>0

Column-1 gives functions and column-2 the nature of the functions.Column-1Column-2A) f:(0,∞)→(0,∞);f(x)=x1+xp) One-one, on toB) f:R−{0}→R,f(x)=x−1xq)One-one but not on toC) f:R−{0}→R,f(x)=x+1xr)On to but not one-oneD) f:R→R,f(x)=2x+sinxs)Neither one-one nor on to

A) f(x)=x1+x⇒f1(x)=1(1+x)2>0 B) f1(x)=1+1x2 C) f1(x)=2+cos>0

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Column-1 gives functions and column-2 the nature of the functions.
Column-1
Column-2
A) $f : (0, \infty) \to (0, \infty); f (x) = \frac{x}{1 + x}$ p) One-one, on to
B) $f : R - {0} \to R, f (x) = x - \frac{1}{x}$ q)One-one but not on to
C) $f : R - {0} \to R, f (x) = x + \frac{1}{x}$ r)On to but not one-one
D) $f : R \to R, f (x) = 2 x + \sin x$ s)Neither one-one nor on to

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$A - q, B - r, C - s, D - p$

$A - q, B - r, C - p, D - s$

$A - q, B - s, C - r, D - p$

$A - q, B - s, C - r, D - r$

answer is D.

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$A) f (x) = \frac{x}{1 + x} \Rightarrow f^{1} (x) = \frac{1}{{(1 + x)}^{2}} > 0 B) f^{1} (x) = 1 + \frac{1}{x^{2}} C) f^{1} (x) = 2 + \cos > 0$

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Column-1	Column-2
A) $f : (0, \infty) \to (0, \infty); f (x) = \frac{x}{1 + x}$	p) One-one, on to
B) $f : R - {0} \to R, f (x) = x - \frac{1}{x}$	q)One-one but not on to
C) $f : R - {0} \to R, f (x) = x + \frac{1}{x}$	r)On to but not one-one
D) $f : R \to R, f (x) = 2 x + \sin x$	s)Neither one-one nor on to

Column-1 gives functions and column-2 the nature of the functions.Column-1Column-2A) f:(0,∞)→(0,∞);f(x)=x1+xp) One-one, on toB) f:R−{0}→R,f(x)=x−1xq)One-one but not on toC) f:R−{0}→R,f(x)=x+1xr)On to but not one-oneD) f:R→R,f(x)=2x+sinxs)Neither one-one nor on to

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