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NDA 1 2024 Maths, question 99

Topic: Algebra

Passage · Q99 to Q100

Let 3f(x)+f(1x)=1x+13f(x)+f\left( \frac{1}{x} \right)=\frac{1}{x}+1

Q99

What is f(x)f(x) equal to ?

  1. (a) 18x−x8+14\frac{1}{8x}-\frac{x}{8}+\frac{1}{4}
  2. (b) 38x−x8+34\frac{3}{8x}-\frac{x}{8}+\frac{3}{4}
  3. (c) 38x+x8+14\frac{3}{8x}+\frac{x}{8}+\frac{1}{4}
  4. (d) 38x−x8+14\frac{3}{8x}-\frac{x}{8}+\frac{1}{4}Correct answer

Solution.

Answer: (d) 38x−x8+14\frac{3}{8x}-\frac{x}{8}+\frac{1}{4}

  1. Replacing x by 1/x gives 3f(1/x) + f(x) = x + 1.
  2. Multiply the original relation by 3 and subtract this new equation: 8f(x) = 3/x − x + 2.
  3. Thus f(x) = 3/(8x) − x/8 + 1/4, with x ≠ 0.