1.Relation and Function
easy

यदि $f(x) = \log \frac{{1 + x}}{{1 - x}}$, तब $f(x)$ है

A

सम फलन

B

$f({x_1})f({x_2}) = f({x_1} + {x_2})$

C

$\frac{{f({x_1})}}{{f({x_2})}} = f({x_1} - {x_2})$

D

विषम फलन

Solution

(d) यहाँ $f(x) = \log \left( {\frac{{1 + x}}{{1 – x}}} \right)$

तथा $f( – x) = \log \left( {\frac{{1 – x}}{{1 + x}}} \right) = \log {\left( {\frac{{1 + x}}{{1 – x}}} \right)^{ – 1}}$

$ = – \log \left( {\frac{{1 + x}}{{1 – x}}} \right) = – f(x)$ ==> $f(x)$ एक विषम फलन है।

Standard 12
Mathematics

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