Algebra – Functions – JEE Main 03 April 2025 Shift 2

Question ID: #1171
JEE Main3 April Shift 2, 2025Algebra

Let $f$ be a function such that $f(x)+3f\left(\frac{24}{x}\right)=4x$, $x\neq 0$. Then $f(3)+f(8)$ is equal to

  • (1) 11
  • (2) 10
  • (3) 12
  • (4) 13

Solution:


$$f(x)+3f\left(\frac{24}{x}\right)=4x$$

$$x = 3$$

$$f(3)+3f\left(\frac{24}{3}\right)=4(3)$$

$$f(3)+3f(8)=12$$

$$x = 8$$

$$f(8)+3f\left(\frac{24}{8}\right)=4(8)$$

$$f(8)+3f(3)=32$$

$$f(3) + 3f(8) + f(8) + 3f(3) = 12 + 32$$

$$4f(3) + 4f(8) = 44$$

$$4(f(3) + f(8)) = 44$$

$$f(3) + f(8) = 11$$

Ans. (1)

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