IIT JEE 1983 Maths - MCQ Question 23

Calculus Level 3

The function $f(x)=\frac {\ln (1+ax) - \ln (1-bx)} x$ is not defined at $x=0$ . The value which should be assigned to $f$ at $x=0$ for constants $a$ and $b$ . So for it to be continuous at $x=0$ , the condition $\text{\_\_\_\_\_\_\_}$ must be fulfilled.

In case you are preparing for IIT JEE, you may want to try IIT JEE 1983 Mathematics Archives .

a-b

None of the others

a+b

\ln a + \ln b

1 solution

Md Zuhair
Feb 15, 2017

Here we need $\lim _{x\to 0} f(x)$ .

Hence So we get $\lim_{x\to 0} \frac {\ln (1+ax) - \ln (1-bx)} x$

As it forms $\dfrac{0}{0}$ form for x=0

We can Apply L'Hospitals Rule

Applying we get,

$\lim_{x\to 0} \dfrac{a}{1+ax} + \dfrac{b}{1-bx}$

So now we can evaluate easily ,

$\lim_{x\to 0} \dfrac{a}{1+ax} + \dfrac{b}{1-bx}$ = $\boxed{a + b}$

IIT JEE 1983 Maths - MCQ Question 23

In case you are preparing for IIT JEE, you may want to try IIT JEE 1983 Mathematics Archives .

1 solution

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