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Asked by g_archanasharma | 01 Apr, 2019, 09:18: AM
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f(x+1) does not exist , because argument of logarithm becomes negative.  Hence option (a) and (b) are not possible.
 
for option (c) , we get product of two log functions, that will not be simplified.
 
Let us test option (d)
 
begin mathsize 12px style f left parenthesis space x subscript 1 space right parenthesis space plus space f left parenthesis space x subscript 2 space right parenthesis space space equals space log open parentheses fraction numerator 1 plus x subscript 1 over denominator 1 minus x subscript 1 end fraction close parentheses space plus space log open parentheses fraction numerator 1 plus x subscript 2 over denominator 1 minus x 2 end fraction close parentheses space equals space log space open square brackets open parentheses fraction numerator 1 plus x subscript 1 over denominator 1 minus x subscript 1 end fraction close parentheses open parentheses fraction numerator 1 plus x subscript 2 over denominator 1 minus x 2 end fraction close parentheses close square brackets space equals space log open parentheses fraction numerator 1 plus space x subscript 1 space plus space x subscript 2 space plus space x subscript 1 x subscript 2 over denominator 1 minus space x subscript 1 space minus space x subscript 2 space plus space x subscript 1 x subscript 2 end fraction close parentheses space end style ...........................(1)
begin mathsize 12px style f open parentheses fraction numerator x subscript 1 space plus x subscript 2 over denominator 1 space plus x subscript 1 x subscript 2 end fraction close parentheses space equals space log open parentheses fraction numerator 1 plus begin display style fraction numerator x subscript 1 space plus space x subscript 2 over denominator 1 space plus x subscript 1 x subscript 2 end fraction end style over denominator 1 minus fraction numerator x subscript 1 space end subscript plus x subscript 2 over denominator 1 space plus x subscript 1 x subscript 2 end fraction end fraction close parentheses space equals space log open parentheses fraction numerator 1 space plus x subscript 1 x subscript 2 space plus space x subscript 1 space plus x subscript 2 over denominator 1 space plus x subscript 1 x subscript 2 space minus space x subscript 1 space minus x subscript 2 end fraction close parentheses end style ....................................(2)
from (1) and (2), it is checked that option (d) is correct
Answered by Thiyagarajan K | 01 Apr, 2019, 12:38: PM

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