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Algebra

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Joined:	31 May 2007
Post:	520

23 Jul 2007 09:02:56 IST

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532

himanshu slove this

None

f:R-->S ,f(x)=sinx-

3cosx+1

function is onto function

then S will be

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naman gupta

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Joined: 26 Mar 2007

Posts: 191

23 Jul 2007 10:13:49 IST

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the ans is S is [-1,3]

see u will have to calculate the max. and min. value of the function
and for "asinx + or - bcosx" system the value lies between -(a²+ b²)^1/2to +-(a²+ b²)^1/2

and hence weget the range to be [ -2 + 1 to 2 + 1]

dhruv bharadwaj

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Joined: 27 Jun 2007

Posts: 68

23 Jul 2007 10:54:03 IST

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or simply,multiply n divide 1st 2 terms by 2 ....
to get

2sin(pi/3 - x) +1

from here,minimum value is:
2*(-1) +1 = -1

n similarly max. is
3

therefore range is [-1,3]
n for onto func,.
co doamin =range...
therefore s=[-1,3]

dhruv bharadwaj

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Joined: 27 Jun 2007

Posts: 68

23 Jul 2007 10:54:38 IST

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sorry m not himanshu....

Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

23 Jul 2007 11:40:21 IST

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When function is onto co-domain is same as range.

So we have to find the range of the function.

f(x) = sinx-

3cosx+1

f(x) = 2sin(

/3 - x) + 1

since -1 <= sin(

/3 - x) <= 1

-2 <= 2sin(

/3 - x) <= 2

-2 + 1 <= 2sin(

/3 - x) + 1 <= 2 + 1

-1 <= 2sin(

/3 - x) + 1 <= 3

Hence range i.e. S is [-1,3]

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