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Differential Calculus

Blazing goIITian

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5 Nov 2008 21:17:07 IST

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If the functions f(x) and g(x) are defined on R -> R such that ...

None

Q is set of rational numbers...

If the functions f(x) and g(x) are defined on R --> R such that

that is, f(x) = 0 if x is rational and x if x isn't rational, and

that is, g(x) = 0 if x is not rational and x if x is rational,

Then ( f - g ) (x) is

(A) one-one and onto

(B) neither one-one nor onto

(C) one-one but not onto

(D) onto but not one-one

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Comments (6)

santhosh in NUS

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5 Nov 2008 21:21:05 IST

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is ans a

Mirka

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5 Nov 2008 21:22:11 IST

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Yup, answer is option (A)...

sriram

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6 Nov 2008 19:52:06 IST

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ans is "a"

Mirka

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6 Nov 2008 20:26:31 IST

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How do u get A??

Can u pls explain?

Rajat Sen

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6 Nov 2008 22:57:22 IST

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suppose $x \subseteq Q$ .

Then $f(x)=0$ but $g(x)=x$

so, $(f-g)(x)=-x$ when $x \subseteq Q$

similarly, $(f-g)(x)=x$ when $x \nsubseteq Q$

moreover rational and irrantional values can never be same and $x$ and $-x$ are

both one-one, so $(f-g)(x)$ is one-one aswell!

moreover it attains all real values, so it is onto!

rate if useful! First Prize

Gaurab Mandal

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7 Nov 2008 08:47:21 IST

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$f(x)=\lbrace0\:;x\,\epsilon\,rational\rbrace\:and\:\lbrace x\:;x\,\not\epsilon\,rational\rbrace$

$g(x)=\lbrace0\:;x\,\not\epsilon\,rational\rbrace\:and\:\lbrace x\:;x\,\epsilon\,rational\rbrace$

$\therefore (f-g)(x)=\lbrace-x\:;x\,\epsilon\,rational\rbrace\:and\:\lbrace x\:;x\,\not\epsilon\,rational\rbrace$

So it attains any Real value: Meaning it is onto.

And we already see from above that it is one-one as rational and irrational values are different......

So correct option is A

So Rajat is correct......please rate as I am correct too......... Winner - Second

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