Consider the function
`f(x)={{:(tankx",",xlt0),(3x+2k^(2)",",xge0):}`
What is the non-zero value of k for which the function is continuous at x=0?

Options

A. `1//4`
B. `1//2`
C. 1
D. `2`

Answer

B

Explanation

Given, `f(x)={{:((tankx)/(x)",",xlt0),(3x+2k^(2)",",xge0):}` ltrbgt When function is continuous at x=0, then
`underset(xto0^(-))limf(x)=underset(xto0^(+))limf(x)=f(0)`
`:.underset(xto0-h)lim((tankx)/(x))=underset(xto0+h)lim(3x+2k^(2))=3(0)+2k^(2)`
`impliesunderset(xto0-h)lim[(tank(0-h))/((0-h))]=underset(xto0+h)lim[3(0+h)+2k^(2)]=2k^(2)`
Therefore, `underset(hto0)lim((tankh)/(h))=2k^(2)`
`k=2k^(2)" "[:'underset(xto0)lim(tanx)/(x)=1]`
Hence, `k=(1)/(2)`