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If $f(x)=log\bigg(\large\frac{1+x}{1-x}\bigg)$, then $f(x)+f(y)$ = ?

$\begin{array}{1 1} f(x+y) \\f(xy) \\f\bigg(\large\frac{x+y}{1+xy}\bigg) \\ f\bigg(\large\frac{x+y}{1-xy}\bigg) \end{array}$

1 Answer

Toolbox:
  • $log(ab)=loga+logb$
Ans- (C)
given: $f(x)=log\bigg(\large\frac{1+x}{1-x}\bigg)$
$f(x)+f(y)=log\bigg(\large\frac{1+x}{1-x}\bigg)+log\bigg(\large\frac{1+y}{1-y}\bigg)$
$=log\bigg(\large\frac{1+x+y+xy}{1-x-y+xy}\bigg)$
Divide num. and deno. by (1+xy)
$=log\bigg(\large\frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}\bigg)$
$=f\bigg(\large\frac{x+y}{1+xy}\bigg)$
answered May 8, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 
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