Sets, Relations & Functions Question - None

Question

Let $$f(2-x)=f(2+x)$$ and $$f(20-x)=f(x)$$ for all $x\in\mathbb{R}$ . If $$f(0)=5$$ , minimum solutions of $$f(x)=5$$ on $$[0,170]$$ :

Accepted Answer

Step 1: Two symmetries → f(t)=f(t-16) → period 16.

Step 2: f(0)=5 → f(16k)=5 for k=0,1,...,10: that's 11 values (0,16,...,160).

Step 3: Symmetry about x=2 gives f(0)=f(4)=5 → f(4+16k)=5 for k=0,...,10: another 11 values (4,20,...,164).

Total minimum = 22.

Answer: (A) 22

Trap: Symmetry about x=2 forces a second AP. Don't count only 16k.

Key Concept: Two reflections → period; symmetry centre shifts by T/2

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