Q247 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
A function $f$ is such that $f(xy) = f(x+y)$ for all real values of $x$ and $y$, and $f(5) = 10$.
What is $f(20) + f(-20)$ equal to?
A function $f$ is such that $f(xy) = f(x+y)$ for all real values of $x$ and $y$, and $f(5) = 10$.
What is $f(20) + f(-20)$ equal to?
- A$0$
- B$10$
- Correct answerC$20$
- D$40$
Solution
Given $f(xy) = f(x+y)$ for all real values of $x$ and $y$.
Substituting $y = 0$, we get $f(x \cdot 0) = f(x + 0)$, which simplifies to $f(0) = f(x)$ for all real $x$.
This implies that $f(x)$ is a constant function for all real values of $x$.
Since $f(5) = 10$, the constant value of the function is $10$. Thus, $f(x) = 10$ for all $x \in \mathbb{R}$.
We need to find the value of $f(20) + f(-20)$.
$f(20) = 10$
$f(-20) = 10$
$f(20) + f(-20) = 10 + 10 = 20$
Answer: $20$
Substituting $y = 0$, we get $f(x \cdot 0) = f(x + 0)$, which simplifies to $f(0) = f(x)$ for all real $x$.
This implies that $f(x)$ is a constant function for all real values of $x$.
Since $f(5) = 10$, the constant value of the function is $10$. Thus, $f(x) = 10$ for all $x \in \mathbb{R}$.
We need to find the value of $f(20) + f(-20)$.
$f(20) = 10$
$f(-20) = 10$
$f(20) + f(-20) = 10 + 10 = 20$
Answer: $20$
Hence, the correct answer is option: 'C'.