Q35 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics
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
- Correct answerA$11$
- B$10$
- C$12$
- D$13$
Solution
$f(x)+3 f\left(\frac{24}{x}\right)=4 x$
Put $x=3 \quad f(3)+3 f(8)=12$
Put $x=8 \quad f(8)+3 f(3)=32$
Add both $\quad 4(\mathrm{f}(3)+\mathrm{f}(8))=44$
$f(3)+f(8)=11$
Put $x=3 \quad f(3)+3 f(8)=12$
Put $x=8 \quad f(8)+3 f(3)=32$
Add both $\quad 4(\mathrm{f}(3)+\mathrm{f}(8))=44$
$f(3)+f(8)=11$
Hence, the correct answer is option: 'A'.