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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Dec 2007 17:33:46 IST
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the function f(x)=p[x+1]+2[x-1] clear[n] is the greatest integer function is continuous at x=1 if
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SHREYA |
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12 Dec 2007 18:01:12 IST
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f(x)= p[x+1] + 2 [ x-1 ]
= p[x] + p + 2[x] - 2 ( Because [x+ a] = [x] + a for all integers a)
=(p+2)[x] + p-2
[x] cannot be continuous at integral values like x=1.
So its coefficient should be 0.
Therefore p+2=0
=> p =-2
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13 Dec 2007 00:35:04 IST
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i 2 got d same ans but my medhod of solving is different
first as given in problem it is continuous at x=1 hence LHL=RHL
hence LHL:Lt f(x)= Lt p[1-h+1]+2[1-h-1]
x-1- h-0 [x=1-h h>0 a small +ve no]
=p-2
n RHL is Lt x-1+ f(x)=Lt h-0 p[1+h+1]+2[1+h+1]=2p
hence as said above LHL=RHL
hence 2p=p-1
p=-2
rate me if im write plz
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13 Dec 2007 00:41:29 IST
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if 2p = p - 2
gives p = -2.
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13 Dec 2007 00:45:18 IST
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hello its printing mistake guys its 2p=p-2 im srry
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