1How can u say that?
Please give explanation?
f(x)= |x2n+1|, n belongs to N then,
A) f(x) is continous but not differentiable at x=0
B) f(x) is differentiable at x=0
C) f(x) is discontinous at x=0
D) none of these.
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4 Answers
1ans:B....
lt x→0+ mod(x^2n+1) = x^(2n+1) = 0
ltx→0- mod(x^2n+1) = -x^(2n+1) = 0
at x=0 f(x)=0....so f(x) is continuous
now if x is +ve, f``(x) = [2n(2n+1)x^2n-1]..so x→0+ it is 0 as n belongs to N
if x is -ve f``(x) = -[2n(2n+1)x^2n-1]..so x→0- it is 0
