if f(x) = x^p cos(1/x) ; x not = 0 & f(0) = 0. where p belongs to integers . then p = ????

Question

p can be - options are
(a) p>0 (b) p<0
(c) p = 0 (d) none of these

Answer 1

x^p=f(x)sec(1/x)
p log x=logf(x)+logsec(1/x)
p=[logf(x)+logsec(1/x)]/logx
p can be >0

Answer 2

(a) p>0. This ensures continuity, since in this case, lim x->0 f(x)=0=f(0). I assume this is what the question is about, even though it should be stated.

If p=0, f(x) oscillates between -1 and 1 more and more rapidly as x->0, so it doesn't even have a limit there.
If p<0, the oscillation becomes infinite, due to the vertical asymptotes of x^p for p<0, as x->0.

Answer 3

that depends ...
you define f(0) = 0
but you say nothing abuot the continuity in this point
so p[ can still be anything

Answer 4

none of the above