JEE Mathematics 2023 Question with Solution

Given: $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$ for $x \in \mathbb{R} - \{0,1\}$.

Find: $f(2)$.

Substitute the values used in the cyclic transformation $x \to \frac{1}{1-x}$.

For $x = 2$,

$$f(2) + f(-1) = 3$$

For $x = -1$,

$$f(-1) + f\left(\frac{1}{2}\right) = 0$$

For $x = \frac{1}{2}$,

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

Now add these three equations:

$$\big(f(2) + f(-1)\big) + \big(f(-1) + f\left(\frac{1}{2}\right)\big) + \big(f\left(\frac{1}{2}\right) + f(2)\big) = 3 + 0 + \frac{3}{2}$$

So,

$$2f(2) + 2f(-1) + 2f\left(\frac{1}{2}\right) = \frac{9}{2}$$

Dividing by $2$,

$$f(2) + f(-1) + f\left(\frac{1}{2}\right) = \frac{9}{4}$$

Using

$$f(-1) + f\left(\frac{1}{2}\right) = 0$$

we get

$$f(2) = \frac{9}{4}$$

Therefore, the correct value is $\frac{9}{4}$.

The solution working gives $f(2) = \frac{9}{4}$, which corresponds to option B. The solution incorrectly labels the correct option as C.

Given: $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$.

Find: $f(2)$.

Notice that

$$2 \xrightarrow{} -1 \xrightarrow{} \frac{1}{2} \xrightarrow{} 2$$

under the map $x \mapsto \frac{1}{1-x}$.

Hence the functional equation generates exactly these three relations:

$$f(2) + f(-1) = 3$$ $$f(-1) + f\left(\frac{1}{2}\right) = 0$$ $$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2}$$

Adding all of them gives

$$2\left[f(2) + f(-1) + f\left(\frac{1}{2}\right)\right] = \frac{9}{2}$$

So,

$$f(2) + f(-1) + f\left(\frac{1}{2}\right) = \frac{9}{4}$$

From the second equation,

$$f(-1) + f\left(\frac{1}{2}\right) = 0$$

Subtracting this from the previous result,

$$f(2) = \frac{9}{4}$$

Therefore, the correct option is B.