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Maths-

General

Easy

Question

If $L straight t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator log begin display style space end style left parenthesis 3 plus x right parenthesis minus log space left parenthesis 3 minus x right parenthesis over denominator x end fraction equals k$ then k=

0
$- \frac{1}{3}$
$\frac{2}{3}$
$\frac{- 2}{3}$

hint Hint:

In the question we will first use L hospital theorem, $fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator f apostrophe left parenthesis x right parenthesis over denominator g apostrophe left parenthesis x right parenthesis end fraction$ , in LHS. Then we will use the value of x to get the value of k.

The correct answer is: $2 over 3$

$L straight t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator log begin display style space end style left parenthesis 3 plus x right parenthesis minus log space left parenthesis 3 minus x right parenthesis over denominator x end fraction equals k$
L.H.S= $L straight t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator log begin display style space end style left parenthesis 3 plus x right parenthesis minus log space left parenthesis 3 minus x right parenthesis over denominator x end fraction$
using L hospital theorem,
$L straight t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator begin display style fraction numerator d blank over denominator d straight x end fraction end style log begin display style space end style left parenthesis 3 plus x right parenthesis minus begin display style fraction numerator d blank over denominator d x end fraction end style log space left parenthesis 3 minus x right parenthesis over denominator begin display style fraction numerator d blank over denominator d x end fraction end style x end fraction equals Lt subscript straight x not stretchy rightwards arrow 0 end subscript fraction numerator begin display style fraction numerator 1 over denominator 3 plus straight x end fraction end style plus begin display style fraction numerator 1 over denominator 3 minus straight x end fraction end style over denominator 1 end fraction equals Lt subscript straight x not stretchy rightwards arrow 0 end subscript fraction numerator 3 minus straight x plus 3 plus straight x over denominator left parenthesis 3 plus straight x right parenthesis left parenthesis 3 minus straight x right parenthesis end fraction equals Lt subscript straight x not stretchy rightwards arrow 0 end subscript fraction numerator 6 over denominator 9 plus straight x squared end fraction equals fraction numerator 6 over denominator 9 plus 0 end fraction equals 6 over 9 equals 2 over 3$
So, k= $2 over 3$

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