MCQ
The function $f (x)=\frac{\log (1+ ax )-\log (1- bx )}{x}$ is not defined at $x=0$. The value which should be assigned to f at $x=0$ so that it is continuous at $x=0$, is
- A$a - b$
- ✓$a+b$
- C$\log a+\log b$
- D$\log a-\log b$
✓
Answer
Correct option: B.
$a+b$
(B)
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\log (1+ ax )-\log (1- bx )}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\frac{ a }{1+ a x}+\frac{ b }{1- b x}}{1}$
$\Rightarrow f (0)= a + b$
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\log (1+ ax )-\log (1- bx )}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\frac{ a }{1+ a x}+\frac{ b }{1- b x}}{1}$
$\Rightarrow f (0)= a + b$
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