Question

If a function $f\left(x\right)$ satisfies $f(2-x)=f(2+x)$ and $f(4-x)=f(4+x)$ for all $x\in R$ and it is given that $\underset{0}{\overset{2}{\int }}f\left(x\right)dx=5,$ then the value of $\underset{0}{\overset{50}{\int }}f\left(x\right)dx$ is equal to

• A. $125$
• B. $\underset{-4}{\overset{46}{\int }}f\left(x\right)dx$
• C. $\underset{1}{\overset{51}{\int }}f\left(x\right)dx$
• D. $\underset{2}{\overset{52}{\int }}f\left(x\right)dx$

Answer 1

Answer: (A), (B)

$\underset{-4}{\overset{46}{\int }}f\left(x\right)dx$

$f(2-x)=f(2+x),f(4-x)=f(4+x)$
Putting $x\to x+2$ in second equation
$\Rightarrow f(2-x)=f(6+x)\Rightarrow f(2+x)=f(6+x)$
$f\left(x\right)$ has period $4.$
Now,
$I=\underset{0}{\overset{50}{\int }}f\left(x\right)dx\Rightarrow I=\underset{0}{\overset{48}{\int }}f\left(x\right)dx+\underset{48}{\overset{50}{\int }}f\left(x\right)dx\Rightarrow I=12\underset{0}{\overset{4}{\int }}f\left(x\right)dx+\underset{0}{\overset{2}{\int }}f\left(x\right)dx[\because f(x)\text{is periodic with period}4,\text{so,}f(x)=f(4-x)=f(4+x\left)\right]\Rightarrow I=24\underset{0}{\overset{2}{\int }}f\left(x\right)dx+5\therefore I=125$

Also,
$\underset{-4}{\overset{46}{\int }}f\left(x\right)dx$
Putting $x\to x+4$, we get
$=\underset{0}{\overset{50}{\int }}f(x+4)dx=\underset{0}{\overset{50}{\int }}f\left(x\right)dx(\because f(x+4)=f(x\left)\right)$
$\underset{1}{\overset{51}{\int }}f\left(x\right)dx$
Putting $x\to x-1$, we get
$=\underset{0}{\overset{50}{\int }}f(x-1)dx\ne \underset{0}{\overset{50}{\int }}f\left(x\right)dx$
$\underset{2}{\overset{52}{\int }}f\left(x\right)dx$
Putting $x\to x-2$, we get
$=\underset{0}{\overset{50}{\int }}f(x-2)dx\ne \underset{0}{\overset{50}{\int }}f\left(x\right)dx$