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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If f(x)+f ⁣(1−1x)=1+xf(x) + f\!\left(1-\dfrac{1}{x}\right) = 1+x for all x∈R−{0,1}x \in \mathbb{R} - \{0,1\},The value of 4f(2)4f(2).
Solution
Answer: 3
Step 1: Set up three equations using substitution
f(x)+f ⁣(x−1x)=1+x⋯(1)f(x) + f\!\left(\frac{x-1}{x}\right) = 1+x \quad \cdots (1)
 Replace x→x−1xx \to \dfrac{x-1}{x} (note 1−xx−1=11−x1-\dfrac{x}{x-1} = \dfrac{1}{1-x}):
f ⁣(x−1x)+f ⁣(11−x)=2x−1x⋯(2)f\!\left(\frac{x-1}{x}\right) + f\!\left(\frac{1}{1-x}\right) = \frac{2x-1}{x} \quad \cdots (2)
 Replace x→11−xx \to \dfrac{1}{1-x} (note 1−(1−x)=x1-(1-x) = x):
f ⁣(11−x)+f(x)=2−x1−x⋯(3)f\!\left(\frac{1}{1-x}\right) + f(x) = \frac{2-x}{1-x} \quad \cdots (3)
 Step 2: Solve for f(x)f(x) (1)+(3)−(2)(1)+(3)-(2):
2f(x)=(1+x)+2−x1−x−2x−1x2f(x) = (1+x) + \frac{2-x}{1-x} - \frac{2x-1}{x}
 Step 3: Evaluate at x=2x = 2
2f(2)=3+0−1−32=3−32=32  ⟹  f(2)=342f(2) = 3 + \frac{0}{-1} - \frac{3}{2} = 3 - \frac{3}{2} = \frac{3}{2} \implies f(2) = \frac{3}{4}
4f(2)=34f(2) = 3
 Answer: 3
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