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Logarithmic Identity

CAT 2002 Slot 1 · Quantitative Ability · Easy · Functions

This is an easy Quantitative Ability question from the CAT 2002 Slot 1 paper. It tests Functions. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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If f(x)=log⁡(1+x1−x)f(x) = \log\left(\frac{1+x}{1-x}\right), then f(x)+f(y)f(x) + f(y) is equal to:

  1. A.

    f(x + y)

  2. B.

    f\left(\frac{x+y}{1+xy}\right)

  3. C.

    (x+y)f\left(\frac{1}{1+xy}\right)

  4. D.

    \frac{f(x)+f(y)}{1+xy}

Answer

B

Explanation

f(x)+f(y)=log⁡(1+x1−x)+log⁡(1+y1−y)=log⁡((1+x)(1+y)(1−x)(1−y))=log⁡(1+x+y+xy1−x−y+xy)f(x) + f(y) = \log\left(\frac{1+x}{1-x}\right) + \log\left(\frac{1+y}{1-y}\right) = \log\left(\frac{(1+x)(1+y)}{(1-x)(1-y)}\right) = \log\left(\frac{1+x+y+xy}{1-x-y+xy}\right). Notice that f(x+y1+xy)=log⁡(1+x+y1+xy1−x+y1+xy)=log⁡(1+xy+x+y1+xy−x−y)f\left(\frac{x+y}{1+xy}\right) = \log\left(\frac{1 + \frac{x+y}{1+xy}}{1 - \frac{x+y}{1+xy}}\right) = \log\left(\frac{1+xy+x+y}{1+xy-x-y}\right). Hence, f(x)+f(y)=f(x+y1+xy)f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right).

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