Functions Question - None

Question

Let f :

\mathbb{R} o \mathbb{R}

be a real valued function such that

f(10 + x) = f(10 - x)

\forall

x \in \mathbb{R}

and

f(20 + x) = -f(20 - x)

\forall

x \in \mathbb{R}

. Then which of the following statements is true -

Accepted Answer

Step 1: Analyze the first condition f(10 + x) = f(10 - x).

This means f is symmetric about x = 10. Substituting x → x - 10: f(x) = f(20 - x).

Step 2: Analyze the second condition f(20 + x) = -f(20 - x).

This means f is anti-symmetric about x = 20. Let y = 20 - x, so x = 20 - y. Then f(40 - y) = -f(y), or equivalently: f(40 - x) = -f(x).

Step 3: Combine the results from Steps 1 and 2.

From Step 1: f(x) = f(20 - x).

From Step 2: f(40 - x) = -f(x).

Replacing x with (20 - x) in Step 2's result: f(40 - (20 - x)) = -f(20 - x), which gives f(20 + x) = -f(20 - x). ✓ (confirms consistency)

Step 4: Find the relationship f(-x) with f(x).

From Step 1 with x = 10: f(10) = f(10), so f is symmetric about x = 10.

From f(x) = f(20 - x) and f(40 - x) = -f(x):

f(40 - x) = -f(20 - (40 - x)) = -f(x - 20).

So: -f(x - 20) = -f(x), giving f(x - 20) = f(x), or f(x) = f(x + 20).

This shows f has period 20.

Step 5: Verify f is odd.

Using f(x) = f(20 - x) and f(40 - x) = -f(x):

f(-x) = f(20 - (-x)) = f(20 + x) = -f(20 - x) = -f(x).

Therefore f(-x) = -f(x), so f is odd.

Step 6: Conclusion.

f(x) is odd (f(-x) = -f(x)) and has period 20 (f(x + 20) = f(x)), so f is both odd and periodic.

∴ Answer: A

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