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Maths-

General

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Question

The value of f (0) so that the function $f open parentheses x close parentheses equals fraction numerator l o g open parentheses 1 plus fraction numerator x over denominator a end fraction close parentheses minus l o g open parentheses 1 minus fraction numerator x over denominator b end fraction close parentheses over denominator x end fraction open parentheses x not equal to 0 close parentheses blank$ is continuous at $x equals 0 blank$ is

$\frac{a + b}{a b}$
$\frac{a - b}{a b}$
$\frac{a b}{a + b}$
$\frac{a b}{a - b}$

hint Hint:

We are given a function. We have to find it's value at f(0). We have to find the value such that the function is continuous at that point. We will use one formula of limit to solve the question.

The correct answer is: $fraction numerator a plus b over denominator a b end fraction$

The given function is $f left parenthesis x right parenthesis space equals space fraction numerator log open parentheses 1 space plus space begin display style x over a end style close parentheses space minus space log open parentheses 1 plus space begin display style x over b end style close parentheses over denominator x end fraction$
We have to find the value of f(0) such that the function is continuous at x = 0
The condition for any function to be continuous at that point is
$L e t space a space b e space a n y space p o i n t f left parenthesis a right parenthesis space equals space limit as x rightwards arrow a of f left parenthesis x right parenthesis$
So, we will solve the above equation using this condition.
$f left parenthesis 0 right parenthesis space equals limit as x rightwards arrow 0 of f left parenthesis x right parenthesis space space space space space space space equals space limit as x rightwards arrow 0 of space fraction numerator log open parentheses 1 space plus begin display style x over a end style close parentheses minus log open parentheses 1 plus begin display style x over b end style close parentheses over denominator x end fraction W e space k n o w space t h a t space space space limit as x rightwards arrow 0 of left parenthesis a space plus space b right parenthesis space equals space limit as x rightwards arrow 0 of a space plus space stack lim b with x rightwards arrow 0 below f left parenthesis 0 right parenthesis space equals space limit as x rightwards arrow 0 of fraction numerator log open parentheses 1 space plus space begin display style x over a end style close parentheses over denominator x end fraction space minus space limit as x space rightwards arrow 0 of fraction numerator log open parentheses 1 space plus space begin display style x over b end style close parentheses over denominator b end fraction T h e r e space i s space f o r m u l a space limit as x rightwards arrow 0 of fraction numerator log open parentheses 1 space space plus space k x close parentheses over denominator x end fraction space equals space k space space space space space space space$
So, the function becomes
$f left parenthesis 0 right parenthesis space equals space 1 over a space minus 1 over b space f left parenthesis 0 right parenthesis space space equals fraction numerator a space minus space b over denominator a b end fraction$
This is the value of f(0) for given function.

For such questions, we should know formula for different limits.

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