JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $f : \mathbb{R} - \{0, 1\} \rightarrow \mathbb{R}$ be a function such that $f (x) + f (\frac{1}{1 - x}) = 1 + x$ Then $f(2)$ is equal to:

Choose the correct answer:

Explanation

Solution

To find $f(2)$ , we need to substitute different values for $x$ into the given equation to create a cycle. The given equation is:

f(x) + f\left(\frac{1}{1 - x}\right) = 1 + x \quad \text{--- (Eq. 1)}

Step 1: Substitute $x = 2$ into (Eq. 1)

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

f(2) + f(-1) = 3 \quad \text{--- (Eq. A)}

Step 2: Substitute $x = -1$ into (Eq. 1)

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

f(-1) + f\left(\frac{1}{2}\right) = 0 \quad \text{--- (Eq. B)}

Step 3: Substitute $x = \frac{1}{2}$ into (Eq. 1)

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

f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \quad \text{--- (Eq. C)}

Step 4: Solve the system of equations

We now have three variables: $f(2)$ , $f(-1)$ , and $f(1/2)$ . Let's add (Eq. A) and (Eq. C), then subtract (Eq. B):

(f(2) + f(-1)) + (f(1/2) + f(2)) - (f(-1) + f(1/2)) = 3 + \frac{3}{2} - 0

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

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

Final Answer:

The value of $f(2)$ is $\frac{9}{4}$ .

The correct option is (3).