If f(x₁)-f(x₂)=f≤ft({x₁-x₂}{1-x₁x₂}) for x₁, x₂ (-1, 1), then what is f(x) equal to?

Correct Answer is : ln⁡(1−x1+x)≤ft(1-x/1+x)ln(1+x1−x)

Correct Answer is : ln⁡(1−x1+x)≤ft(1-x/1+x)ln(1+x1−x)

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UPSC NDA I 2016 Mathematics Solved Paper

Question : 5 of 120

Marks: +1, -0

f(x_{1})-f(x_{2})=f\left(\frac{x_{1}-x_{2}}{1-x_{1}x_{2}}\right)

x_{1}, x_{2} \in (-1, 1)

f(x)

equal to?

Solution:

Concept:

Find

f(0)

by substituting

x_1 = x_2 = 0

into the given equation, then check which option gives

f(0)=0

Explanation:

Given:

f(x_1) - f(x_2) = f\left(\frac{x_1 - x_2}{1 - x_1 x_2}\right)

for

x_1, x_2 \in (-1, 1)

Put

x_1 = x_2 = 0

f(0) - f(0) = f(0)

gives

0 = f(0)

. So

f(0) = 0

Now evaluate each option at

x = 0

Option A:

f(x) = \ln\left(\frac{1-x}{1+x}\right)

gives

f(0) = \ln(1) = 0

Option B:

f(x) = \ln\left(\frac{2+x}{1-x}\right)

gives

f(0) = \ln(2) \neq 0

Option C:

f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right)

gives

f(0) = \tan^{-1}(1) = \frac{\pi}{4} \neq 0

Option D:

f(x) = \tan^{-1}\left(\frac{1+x}{1-x}\right)

gives

f(0) = \tan^{-1}(1) = \frac{\pi}{4} \neq 0

Only option A satisfies

f(0) = 0

Answer:

Option A:

\ln\left(\frac{1-x}{1+x}\right)

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