Let f be a function such that f(x)+3 f (24/x )=4 x , x neq 0 . Then f(3)+f(8) is equal to

Mathematics · Functions

JEE Main 2025 — 3 April, Evening Shift — Question 31

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

$f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0 …(1)$

replace $x$ by $\frac{24}{x}$

$f\left(\frac{24}{x}\right)+3 f\left(\frac{24}{\frac{24}{x}}\right)=4\left(\frac{24}{x}\right)=\frac{96}{x} … (2)$

$3 \times(2)-(1)$

$\Rightarrow 8 f(x)=\frac{96.3}{x}-4 x \Rightarrow f(x)=\frac{36}{x}-\frac{x}{2}$

$f(3)+f(8)=\left(12-\frac{3}{2}\right)+\left(\frac{36}{8}-4\right)$

$=8+\frac{36}{8}-\frac{12}{8}=11$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.