The left hand derivative of `f(x)=[x]sin(pix)` at `x = k`, `k in Z`, is

Options

A. `(-1)^(k) (k-1)pi`
B. `(-1)^(k-1) (k-1)pi`
C. `(-1)^(k)kpi`
D. `(-1)^(k-1)kpi`

Answer

A

Explanation

We have
`f(x)=[x] sin pi x`
`therefore ("LHD at x=k")=underset(x to k^(-))lim (f(x)-f(k))/(x-k)`
`Rightarrow ("LHD at x=k")=underset(h to 0)lim (f(k-h)-f(k))/(-h)`
`Rightarrow ("LHD at x=k")=underset(h to 0)lim ([k-h)sin (pik-pih)-[k]sin pik)/(-h)`
`Rightarrow ("LHD at x=k")=underset(h to 0)lim ((-1)^(k-1)[k-h]sin pi h-0)/(-h)`
`Rightarrow ("LHD at x=k")=underset(h to 0)lim (-1)^(k) [k-h] (sin pi h)/(h)`
`Rightarrow ("LHD at x=k")=(-1)^(k) (k-1) underset(h to 0)lim (sin pi h)/(h)=(-1)^(k) (k-1)pi`