JEE Mathematics 2025 Question with Solution

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

Find: $f(3) + f(8)$

Substitute $x = 3$ in the given equation:

$$f(3) + 3f\left(\frac{24}{3}\right) = 4(3)$$ $$f(3) + 3f(8) = 12$$

Substitute $x = 8$ in the given equation:

$$f(8) + 3f\left(\frac{24}{8}\right) = 4(8)$$ $$f(8) + 3f(3) = 32$$

Now add the two equations:

$$\left(f(3) + 3f(8)\right) + \left(f(8) + 3f(3)\right) = 12 + 32$$ $$4f(3) + 4f(8) = 44$$

Divide by $4$:

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

Therefore, the correct option is A and the value of $f(3) + f(8)$ is $11$.

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

Find: $f(3) + f(8)$

The key observation is that $3$ and $8$ are paired by the transformation $x \mapsto \frac{24}{x}$ because $\frac{24}{3} = 8$ and $\frac{24}{8} = 3$.

So, substitute the two connected values directly:

1. For $x = 3$,

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

2. For $x = 8$,

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

Add them to eliminate the asymmetric coefficients:

$$\begin{aligned} &f(3) + 3f(8) + f(8) + 3f(3) = 12 + 32 \\ &4f(3) + 4f(8) = 44 \\ &4\bigl(f(3) + f(8)\bigr) = 44 \end{aligned}$$

Hence,

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

Therefore, the required sum is $11$, so the correct option is A.

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

Find: $f(3) + f(8)$

Because $3$ and $8$ map into each other under $\frac{24}{x}$, write the equation once at $x=3$ and once at $x=8$:

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

A quick shortcut is to add them immediately, since both unknowns then get coefficient $4$:

$$4f(3) + 4f(8) = 44$$ $$f(3) + f(8) = 11$$

Therefore, the correct option is A.