Question

Let $f:R\to R$ be a 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 $10$

$f(2-x)=f(2+x)$

$x\to x=2$

$f(4-x)=f\left(x\right)---\left(1\right)$

f(4-x)=f(4+x)$

$x\to x-4$

$f(8-x)=f\left(x\right)---\left(2\right)$

$\Rightarrow f(4-x)=f(8-4)$

$4-x\to x$

$f\left(x\right)=f(x+4)---\left(3\right)$

period of $f=4$

${\int }_0^{2}f\left(x\right)=0.5$

$\Rightarrow {\int }_0^{2}f\left(x\right)={\int }_0^{2}f(2-x)={\int }_0^{2}f(2+x)$

$={\int }_2^{4}f\left(t\right)$

${\int }_2^{4}f\left(x\right)=0.5$

$\therefore {\int }_0^{2}f\left(x\right)={\int }_0^{2}f\left(x\right)+{\int }_2^{4}f\left(x\right)=0.5+0.5=1$

$\therefore$ Area of required region

$=\frac{50-10}{4}\times 1=10units$