Find the no of solutions.
2cosx=\left|sinx \right|x belongs to 0 to 2pie
1at pi / 2 , sin x is 1 ( seems you dozed off ) anyway , answer is 3
1if two curves are symmetric abt any fixed line doesn't make any thing easier -- you can't just say what you said -- tell you what , this came in FIITJEE and they also solved it wrongly
1As everyone knows...the 2 bumps r symmetric about x = pi.
And the graph of 2^cos x is also symmetric about x = pi.
So wht u've predicted is merely a wrong assumption.
And soumya stop thinking abt. this ... the Joker has already said "why so serious ??"
1but once the graph of 2^cos x cuts that of | sin x | , the following things happen --
as | sin x | and cos x are both increasing in the interval pi - 3 pi / 2 , so obviously the graph of 2 ^ cos x will rise more sharply than that of the sine one -- so I don't think there will be 4 solutions ---
maybe I really am wrong -- I welcome everyone to clear this mess up -- : ) : 0
1Once again I say.it's 4 solutions...........isn't it Joker ???
1so is the DARK KNIGHT not once again teaming up with the JOKER ???? the answer is 4 indeed.
1I really dozed off..........like years of rivalry once again the DARK KNIGHT and the JOKER differ in their viewpoints ....... the answer is 3 indeed.....now it's too tedious to post the graphs....but i've done it ... kk thnx dude...'twas a good sum. neways...2^cos x graph was right na ??
6its easy ...nishant bhaiya u gave d question to soumya...n he gave me....
1ARE YOU MAD ???????????????? HOW CAN | sin x | HAVE A VALUE 0 AT pi / 2 ????????????????
1This is the first time ever that the Joker and the Dark Knight have shared a common idea .
1Clearly der r 4 solns. .
the graph of 2^cos x cuts the sin curve in the middle 2 bumps at 4 points.
1hey kalyan , seems no body is interested ---- draw the graphs folks
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