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@Apu
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Limits aa gyi , can't figure out the lcd part
3cos(x) ranges from [-3,3] in the given interval, if 3cos(x)<log(3), the g.i.f. turns zero. For log(3)<3cos(3)<log(3)+log(2)=log(6), g.i.f. turns 1, and so on.
e^3 is approximately (2.7)^3 ig which is 243*81/10^3 which is 19.683
Toh this the maximum the g.i.f can reach is 6.
So it goes from 6 to 5 (1 disc.), 5 to 4 (2nd disc.) and so on till it reaches zero and remains zero for the rest.
So 6 discontinuities?
ans shi hai but my braincells are turning off
so subah dekhunga
@vj25_ tune bhi aise hi kia tha ??
yes...function aa gya tha but discontinuity nahi nikli thi 😢
same
i can just ignore the inequalities and directly analyse the extremes of the function
e^3 is nearly 19.6 and e^-3 = 0.04 . so 6 multiples of 3 are covered and thus 6 breaks will be seen
Sure but note that if we had sine function or smth we would have had to think about where it achieves maximum or is increasing.
follows a similar pattern ??
Yep sin and cos follow similar patterns but I'm just sayin'. Its not necessary that just checking end points is not safe every time.
alrr
got it
thanx
+solved @SirLancelotDuLac
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