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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Sep 2008 13:17:07 IST
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2^cosx = |sin x| within the interval (0,2pi)
Ans : 2;
How to solve....................:(
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21 Sep 2008 14:03:26 IST
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well you can easily try it by observation
as x =  0r 
BEST OF LUCk
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21 Sep 2008 14:07:45 IST
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can anyone tell the systematic way of solving the problem!!!!?????
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21 Sep 2008 14:43:22 IST
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min value of 2^cosx=1 max value of sinx=1 therefore solution is one as mod is given 2 solutions
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21 Sep 2008 14:56:29 IST
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NO, man you are wrong !!!!!!!!
min value of  is not 0 as cosx can be -ve also
cosx (-1,1) so min value = 1/2
best of luck
plz somebody give the correct ans with systematic soln
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21 Sep 2008 15:13:47 IST
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2^cosx = |sin x| within the interval (0,2pi)
Just draw the graphs of both expressions
2^cosx (and cos varies from -1 to 1, so draw accordingly)
And |sinx| , draw the sin graph and invert everything to positive y axis. Then you'll easily see the number of intersections between the 2 graphs which is equal to the number of solutions.
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- Gaurav Ragtah (spideyunlimited)
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21 Sep 2008 15:19:32 IST
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Acc. to graph, there are 4 solutions! not 2.
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- Gaurav Ragtah (spideyunlimited)
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21 Sep 2008 15:45:44 IST
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right abt the minimum value of 2^cosx........ its not zero
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21 Sep 2008 15:53:39 IST
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hey bro @ SPIDY ........ CAN U PLEASE depict the graph in kind o detail ?
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21 Sep 2008 15:58:54 IST
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Least value of 2^ cosx = 1/2 and max value is 2.
Click on the graph to view in detail.
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- Gaurav Ragtah (spideyunlimited)
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21 Sep 2008 16:06:43 IST
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hey bro @ SPIDY ........ CAN U PLEASE depict the graph in kind o detail ?
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21 Sep 2008 16:57:14 IST
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since cos varies from -1 to 1, so 2^cosx varies from 1/2 to 2.
And sin x graph , i think you know. So mod of that graph will have everything reflected above x axis.
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- Gaurav Ragtah (spideyunlimited)
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22 Sep 2008 11:00:37 IST
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graph is absolutely correct.So the no. of solutions in given interval is 4."This is the right answer .
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***T.Venkat*** |
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