Question

Let $f:R\to R$ be function such that $f(2-x)=f(2+x)$ and $f(4-x)=f(4+x)\forall x\in R$ and ${\int }_0^{2}f\left(x\right)dx=0.5$, then area of region bounded by $y=f\left(x\right)$ and lines $x=10$ and $x=50$ is ____

• A. $10$
• B. $20$
• C. $50$
• D. $100$

Answer 1

The correct option is A $10$

$f(2-x)=f(2+x);f(4-x)=f(4+x);f\left(x\right)=f(4-x);$ periodicity is 4.

${\int }_0^{2}f\left(x\right)dx={\int }_0^{2}f(2-x)dx={\int }_0^{2}f(2+x)dx={\int }_2^{4}f\left(x\right)dx⟹{\int }_0^{4}f\left(x\right)dx=1$ so the same area repeats between every 4 units.Area between that 4 units is symmetric so it can be said that $\frac{50-10}{4}=10$