Question

If f \((5+x) = f (5-x)\) for every real x, and \(f(x)=0\) has four distinct real roots, then the sum of these roots is

• A. 0
• B. 40
• C. 10
• D. 20

Answer 1

The Correct Option is D

To solve the problem, we start with the given property of the function:

\(f(5+x) = f(5-x)\) for every real \(x\).

This implies that the function is symmetric about the line \(x = 5\). Therefore, if \(a\) is a root, then \(10-a\) is also a root due to the symmetry.

We are given that \(f(x)=0\) has four distinct real roots. Let's denote these roots as \(r_1, r_2, r_3, r_4\). Due to the symmetry, they can be paired as follows:

• \(r_1 + (10 - r_1) = 10\)
• \(r_2 + (10 - r_2) = 10\)

Thus, each pair of symmetric roots sums to 10. Therefore, the total sum of all roots is:

\(10 + 10 = 20\).

Thus, the sum of the four distinct real roots is 20.

Answer 2

Let r be a root of the function, so: 
\( f(r) = 0 \)

Using the functional equation given: 
\( f(5 - x) = f(5 + x) \)

We substitute: 
\( f(r) = f(5 - (5 - r)) = f(5 + (5 - r)) = f(10 - r) \)

Hence, if \( r \) is a root, then so is \( 10 - r \). 
This means roots occur in symmetric pairs about \( x = 5 \).

Example:

If there are four distinct roots: 
\( r_1, (10 - r_1), r_2, (10 - r_2) \)

Then, the sum of roots is: 
\( r_1 + (10 - r_1) + r_2 + (10 - r_2) = 10 + 10 = 20 \)

 Final Answer:

Option (D): The sum of the four different roots is 20.