Question

Let a function $f(x)$ satisfy \[ 3f(x)+2f\!\left(\frac{m}{19x}\right)=5x \] where $m=\sum_{i=1}^{9} i^2$. Find $f(5)+f(2)$.

Hint:

For functional equations, always try replacing $x$ by reciprocal expressions to form solvable systems.

Answer 1

Correct Answer: 0

Step 1: Evaluate the value of m

m = ∑_i=1⁹ i²

m = (9 × 10 × 19) / 6

m = 285

---

Step 2: Write the given functional equation

3f(x) + 2f(285 / 19x) = 5x

Since 285 / 19 = 15,

3f(x) + 2f(15 / x) = 5x  ……(1)

---

Step 3: Replace x by (15 / x)

3f(15 / x) + 2f(x) = 5(15 / x)

3f(15 / x) + 2f(x) = 75 / x  ……(2)

---

Step 4: Solve equations (1) and (2)

Multiply (1) by 2 and (2) by 3:

6f(x) + 4f(15 / x) = 10x  ……(3)

6f(15 / x) + 4f(x) = 225 / x  ……(4)

Subtract (4) from (3):

2f(x) − 2f(15 / x) = 10x − 225 / x

f(x) − f(15 / x) = (5x − 112.5 / x)

---

Step 5: Solve for f(x)

From equation (1):

3f(x) = 5x − 2f(15 / x)

Substitute f(15 / x) = f(x) − (5x − 112.5 / x)

3f(x) = 5x − 2[f(x) − 5x + 112.5 / x]

3f(x) = 5x − 2f(x) + 10x − 225 / x

5f(x) = 15x − 225 / x

f(x) = 3x − 45 / x

---

Step 6: Compute required values

f(5) = 3(5) − 45/5 = 15 − 9 = 6

f(2) = 3(2) − 45/2 = 6 − 22.5 = −16.5

---

Final Answer:

f(5) + f(2) = 6 − 16.5 = −10.5