|Nov15-05, 09:51 AM||#1|
integral of f(t)dt from 0 to x tends to 5 as |x|->1.Find values of 'a' for which g(x)=2x + (integral of f(t)dt from 0 to x) = a has atlest 2 roots of opp.signs. in the interval (-1,1)..
I think that since there shud be atlest 2 roots,g(x) shud cut/touch the x-axis atleast twice..whichc means that the min value of g(x) can be either 0 or some -ve value..(assuming it is differentiable at all points in itz domain)..for the latter case...f(0)<0 ...am stuck now...please help
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