Functions & Logarithms PYQ | JEE Main 2025 | Smart Identity Trick

 

❓ Question

Given

$$f(x)=\log\left(\frac{1+x}{1-x}\right),$$

find

$$f\left(\frac{3x+x^3}{1+3x^2}\right).$$

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✍️ Solution

Let

$$y=\frac{3x+x^3}{1+3x^2}.$$

Then,

$$f(y)=\log\left(\frac{1+y}{1-y}\right).$$

Substituting the value of $y$,

$$f(y) = \log\left( \frac{1+\dfrac{3x+x^3}{1+3x^2}} {1-\dfrac{3x+x^3}{1+3x^2}} \right).$$

Combining the fractions,

$$= \log\left( \frac{1+3x^2+3x+x^3} {1+3x^2-3x-x^3} \right).$$

Now,

$$1+3x+3x^2+x^3=(1+x)^3,$$

and

$$1-3x+3x^2-x^3=(1-x)^3.$$

Hence,

$$f(y) = \log\left( \frac{(1+x)^3}{(1-x)^3} \right).$$

Using

$$\log(a^n)=n\log a,$$

we get

$$f(y) = 3\log\left(\frac{1+x}{1-x}\right).$$

Since

$$f(x)=\log\left(\frac{1+x}{1-x}\right),$$

therefore,

$$\boxed{f(y)=3f(x).}$$

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✅ Final Answer

$$\boxed{f\left(\frac{3x+x^3}{1+3x^2}\right)=3f(x)}$$