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Differential Calculus

Blazing goIITian

Joined:	5 Jun 2007
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16 Apr 2009 12:31:07 IST

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769

bijection or not??

Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Differential Calculus

Let f:R->R be given by f(x) = (x+1)² - 1 , x-1. Find inverse of x.

a) f^-1(x) = -1 +

b) f^-1(x) = -1 -

c) f^-1(x) = -1

d) f^-1(x) =

The answer given is (a) which is indeed the answer if f is a bijection.

But how is f bijection??

injectivity can be proved. But surjectivity of 'f' is not possible.

Putting y = f(x) we get

x = -1 + . and so for y-1, x doesn't exist. But y-1 is in the codomain.

So Range Codomain.

Hence it is not surjecton. So f is not invertible.

Please HELP

Comments (15)

Killer Kiran

Blazing goIITian

Joined: 6 Aug 2007

Posts: 1860

16 Apr 2009 13:37:56 IST

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since x is greater than 1...it is bijective....

Swapnika.... finally into a branded college....NIT m cuming..!!!

Blazing goIITian

Joined: 7 Aug 2008

Posts: 1047

16 Apr 2009 13:39:23 IST

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it shud be bijective

NugoRama

Blazing goIITian

Joined: 11 Mar 2009

Posts: 5564

16 Apr 2009 13:40:50 IST

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totally agreed biki....it is not invertible in R--->R ..

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

16 Apr 2009 14:15:06 IST

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for x -1

all values of y dont satisfy...

Here the range of the function for x -1 is y -1. But the codomain is R.

So RangeCodomain.

So it is not onto function and hence cannot be inverted...

For function to be invertible, the codomain should have been [ -1, ).

Am i not correct???

NugoRama

Blazing goIITian

Joined: 11 Mar 2009

Posts: 5564

16 Apr 2009 14:17:01 IST

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yes ..function is only invertible if codomain is [-1,infinity)..totally agreed ..thats wat i m saying ..function not invertible

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

16 Apr 2009 14:34:24 IST

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for x -1

all values of y dont satisfy...

Here the range of the function for x -1 is y -1. But the codomain is R.

So RangeCodomain.

So it is not onto function and hence cannot be inverted...

For function to be invertible, the codomain should have been [ -1, ).

Am i not correct???

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

18 Apr 2009 23:27:56 IST

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please tell... is the question wrong or not... the graph shows it is..

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

19 Apr 2009 10:38:48 IST

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please

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

19 Apr 2009 22:28:54 IST

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is this question so bad??

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

20 Apr 2009 13:27:26 IST

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friend...they themselves hav said... "if it is bijective"....u must take it as bijective....their objective is to make urself calculate the inverse....n thts it...

prudhviraj

Scorching goIITian

Joined: 23 Mar 2009

Posts: 293

20 Apr 2009 14:40:04 IST

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domain shud be mentioned otherwise assume it as bijective

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

20 Apr 2009 19:16:21 IST

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what are u saying... i cant get it brother...

they clearly mentioned the domain and codomain to be R.

But we can clearly see that the function is bijective if and only if the codomain is [ -1, )

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

20 Apr 2009 19:17:48 IST

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why should we assume ???

what are u saying... i cant get it brother...

they clearly mentioned the domain and codomain to be R.

But we can clearly see that the function is bijective if and only if the codomain is [ -1, )

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

21 Apr 2009 11:10:50 IST

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ya...true to maths its a wrong ques. but why u r taking so much tension over tht thing.... they just want to know whether u know how to find inverse or not...well we r getting free salutes :D :)

biki ....

Blazing goIITian

Joined: 5 Jun 2007

Posts: 978

21 May 2009 08:25:31 IST

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this question came in aieee 09... gr8 ...

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