Let q(x)= cases (1+kx) x , & x 0, 2, & x=0

MATHEMATICS · Single correct · CBSE · Mathematics · Continuity and Differentiability · Continuity and Differentiability

Problem

Let

q(x)= \begin{cases} \displaystyle\frac{\ln(1+kx)}{x}, & x\ne 0,\\ 2, & x=0. \end{cases}

For which value of

k

is

q

continuous at

x=0

? (a)

1

(b)

0

(c)

2

(d)

e

Related theory

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