If f(x)= (1+ax)- (1-bx) x ,&x 0 ,&x=0 and f(x) is continous at x … - Vidyadip

MCQ

If $\text{f(x)}=\begin{cases}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ and f(x) is continous at x = 0, then the value of k is:

A
a - b
✓
a + b
C
$\log\text{a}+\log\text{b}$
D
none of these

✓

Answer

Correct option: B.

a + b

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$

$\text{a}+\text{b}=\text{k}$

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