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

The correct answer is (D): \(20\)

Given \(f (5+x) = f (5-x)\)

Put \(x = x-5\)

\(f(x) = f(10-x)\)

Let a, b be two roots of \(f(x)=0\), then \(10-a,10-b\) are also roots of \(f (x) = 0\)

\(∴\) Hence sum of the roots = \(a+b+10-a+10-b=20\)

Answer 2

Let 'r' represent the function's root. Consequently, \(f(r) = 0\). We can represent this as.
\(f(r) = f(5 - (5 - r))\)
Given the relationship:
\(f(5−x)=f(5+x); 𝑓 ( 𝑟 ) = 𝑓 { 5 − ( 5 − 𝑟 ) } = 𝑓 { 5 + ( 5 − 𝑟 ) } f(r)\)
\(=f\{5−(5−r)\}=f\{5+(5−r)\} \)
\(∴ 𝑓 ( 𝑟 ) = 𝑓 ( 10 − 𝑟 ) \)
As a result, each root 'r' is linked to a corresponding root \('(10-r)';\) together, these form a pair. 
Let's assume the following for even different roots, in this case four: 
\(r_1, (10 - r_1), r_2, (10 - r_2)\)
The total of these roots is equal to =\(r_1 + (10 - r_1) + r_2 + (10 - r_2) = 20\)
 Therefore, Option D is the right response.