Question

If $f(2-x)=f(2+x)$ and $f(4-x)=f(4+x)$ for all x and f(x) is a function for which ${\int }_0^{2}f\left(x\right)dx=5$, then ${\int }_0^{50}f\left(x\right)dx$ is equal to

• A. $125$
• B. ${\int }_{-4}^{46}f\left(x\right)dx$
• C. ${\int }_1^{51}f\left(x\right)dx$
• D. ${\int }_2^{52}f\left(x\right)dx$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $125$
B ${\int }_{-4}^{46}f\left(x\right)dx$
D ${\int }_2^{52}f\left(x\right)dx$

$f(2-x)=f(2+x),f(4-x)=f(4+x)$ or $f(4+x)=f(4-x)=f(2+2-x)=f(2-(2-x\left)\right)=f\left(x\right)$
Thus, the period of f(x) is 4.
${\int }_0^{50}f\left(x\right)dx={\int }_0^{48}f\left(x\right)dx+{\int }_{48}^{50}f\left(x\right)dx=12{\int }_0^{4}f\left(x\right)dx+{\int }_0^{2}f\left(x\right)dx$
[In second integral, replacing x by x + 48 and then using $f\left(x\right)=f(x+48)]=12({\int }_0^{2}f\left(x\right)dx+{\int }_0^{2}f(4-x)dx)+5$
$=12\left({\int }_0^{2}f\right(x)dx+{\int }_0^{2}f(4+x\left)dx\right)+5=24{\int }_0^{2}f\left(x\right)dx+5=125$
${\int }_{-4}^{46}f\left(x\right)dx={\int }_{-4}^{-2}f\left(x\right)dx+{\int }_{-2}^{-2+48}f\left(x\right)dx={\int }_0^{2}f(x+4)dx+12{\int }_0^{4}f\left(x\right)dx$
$={\int }_0^{2}f\left(x\right)dx+24{\int }_0^{2}f\left(x\right)dx=125$
Also, ${\int }_2^{52}f\left(x\right)dx={\int }_2^{4}f\left(x\right)dx+{\int }_4^{4+48}f\left(x\right)dx={\int }_0^{2}f(4-x)dx+12{\int }_0^{4}f\left(x\right)dx$
$={\int }_0^{2}f(4+x)dx+24{\int }_0^{2}f\left(x\right)dx={\int }_0^{2}f\left(x\right)dx+24{\int }_0^{2}f\left(x\right)dx=125$
${\int }_1^{51}f\left(x\right)dx={\int }_1^{3}f\left(x\right)dx+{\int }_3^{3+48}f\left(x\right)dx={\int }_1^{3}f\left(x\right)dx+12{\int }_0^{4}f\left(x\right)dx$
$={\int }_0^{2}f(x+1)dx+24{\int }_0^{2}f\left(x\right)dx\ne 125$