Let `f : R -> R` is a function satisfying `f(2-x) = f(2 + x) and f(20-x)=f(x),AA x in R`. On the basis of above information, answer the following questions If `f(0)=5`, then minimum possible number of values of x satisfying `f(x) = 5`, for `x in [10, 170]` is

Options

A. 21
B. 12
C. 11
D. 22

Explanation

To solve the problem, we will analyze the given functional equations and derive the necessary conclusions step by step.

### Step 1: Analyze the first functional equation
We start with the equation:
\[ f(2 - x) = f(2 + x) \]
This indicates that the function \( f \) is symmetric about \( x = 2 \).

### Step 2: Substitute \( x \) with \( 2 - x \)
By substituting \( x \) with \( 2 - x \) in the first equation:
\[ f(2 - (2 - x)) = f(2 + (2 - x)) \]
This simplifies to:
\[ f(x) = f(4 - x) \]
Let’s label this as Equation (2).

### Step 3: Analyze the second functional equation
The second equation given is:
\[ f(20 - x) = f(x) \]
This shows that the function \( f \) is symmetric about \( x = 10 \).

### Step 4: Combine the results
From Equation (2) and the second functional equation, we can derive:
\[ f(4 - x) = f(x) \]
and
\[ f(20 - x) = f(x) \]

### Step 5: Substitute \( x \) in the second equation
Now, substitute \( x \) with \( 4 - x \) in the second equation:
\[ f(20 - (4 - x)) = f(4 - x) \]
This simplifies to:
\[ f(16 + x) = f(4 - x) \]
Since \( f(4 - x) = f(x) \), we have:
\[ f(16 + x) = f(x) \]

### Step 6: Establish periodicity
From the above, we can conclude that the function \( f \) is periodic with a period of \( 16 \):
\[ f(x + 16) = f(x) \]

### Step 7: Use the periodicity
Given that \( f(0) = 5 \), we can find:
\[ f(16) = f(0) = 5 \]
\[ f(32) = f(0) = 5 \]
Continuing this pattern, we find:
\[ f(48) = f(0) = 5 \]
\[ f(64) = f(0) = 5 \]
\[ f(80) = f(0) = 5 \]
\[ f(96) = f(0) = 5 \]
\[ f(112) = f(0) = 5 \]
\[ f(128) = f(0) = 5 \]
\[ f(144) = f(0) = 5 \]
\[ f(160) = f(0) = 5 \]
\[ f(176) = f(0) = 5 \]

### Step 8: Count the values in the range
Now we need to count how many of these values fall within the interval \( [10, 170] \):
- The values of \( x \) that satisfy \( f(x) = 5 \) in the interval \( [10, 170] \) are:
- \( 16, 32, 48, 64, 80, 96, 112, 128, 144, 160 \).

### Conclusion
Thus, the minimum possible number of values of \( x \) satisfying \( f(x) = 5 \) for \( x \in [10, 170] \) is:
**10**