If `f(x) = underset(r = 1)overset(n)(sum)a_(r)|x|^(r)`, where `a_(i)` s are real constants, then f(x) is

Options

A. continuous at x=0 for all `a_(1)`
B. differentiable at x=0 for all `a_(i) in R`
C. differentiable at x=0 for all `a_(2k+1)=0`
D. none of these

Answer

A, C

Explanation

We know that `|x|^(r),r=0,1,2....` are all continuous everywhere.
`f(x)=underset(r=1)overset(n)sum a_(r)|x|^(r)` is everywhere continuous
`"Since", |x|, |x|^(3),|x|^(5),...etc` are not differentiable at x=0
`f(x)=underset(r=1)overset(n)sum a_(r)|x|^(r)` is nor differentiable at x=0, if any one of `a_(1),a_(3),a_(5)...` is non-zero.
Thus, for f(x) is to be differentiable at x=0, we must have
`a_(1)=a_(3)=a_(5).....0 i.e. a_(2k+1)=0`
Hence, options (a) and (c) are correct,