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Question

If the functions f and g are defined on RR such that
f(x)={0, if x is rationalx, if x is irrational and
g(x)={0, if x is irrationalx, if x is rational,
then (fg)(x) is

A
a bijective function
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B
neither a one-one nor an onto function
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C
a one-one but not an onto function
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D
an onto but not a one-one function
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Solution

The correct option is A a bijective function
(fg)(x)={x, if x is rationalx, if x is irrational

Let x1,x2 rational (fg)(x1)=(fg)(x2)x1=x2x1=x2 (i)

Let x1,x2 irrational
(fg)(x1)=(fg)(x2)x1=x2 (ii)

Let x1 irrational, x2 rational
(fg)(x1)=(fg)(x2)x1=x2x+x2=0(iii)
This is not possible as sum of rational and irrational cannot be 0.
So, this is a contradiction to assumption x1 irrational, x2 rational
From equations (i),(ii) and (iii), we get
(fg)(x) is one-one function.

Let y rational
y=(fg)(x)y=x
x=y rational (iv)

y Irrational
y=(fg)(x)y=x
x irrational (v)

from (iv) and (v) (fg)(x) is onto function.
Hence (fg)(x) is a bijective function.


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