lets
f(x) = number of solutions for (sin y = 1) where y is [0,x];
g(x) = number of solutions for (sin^2y = 1) where y is [0-x];
c(x) = f(x)/g(x);
find the value of c(x) lim 1/x->0;
answer is simple and to the point
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the ans shud be 1/2,,,,,,,
ReplyDeleteas 1/x->0;
x->infinte....
domain 0 to infinite,,,,,
comparin da graphs of both da functions,,,
sin^2x has twice as many sol. than sinx,,,
perfect answer by pankaj
ReplyDeleteit has twice as many solutions as sin x because in the graph it intersect double the time by graph for a fixed period integral multiple of their combined period
as sin crest trough is converted into crest after squaring it