Question

If, $A_1=\underset{n}{\overset{n+1}{\int }}(min\{|x-n|,|x-(n+1)|\left\}\right)dx$,
$A_2=\underset{n+1}{\overset{n+2}{\int }}(|x-n|-|x-(n+1\left)|\right)dx{A}_3=\underset{n+2}{\overset{n+3}{\int }}(|x-(n+4)|-|x-(n+3\left)|\right)dx$ and $g\left(x\right)=A_1+A_2+A_3,$ where $n\in N,$ then

• A. $A_1+A_2+A_3=9$
• B. $A_1+A_2+A_3=\frac{9}{4}$
• C. $\sum _{n=1}^{100}g\left(x\right)=\frac{900}{4}$
• D. $\sum _{n=1}^{100}g\left(x\right)=300$

Answer 1

Answer: (B), (C)

$\sum _{n=1}^{100}g\left(x\right)=\frac{900}{4}$

Here, min $\{|x-n|,|x-(n+1\left)|\right\}$ can be shown as
$\therefore A_1=\underset{n}{\overset{n+1}{\int }}(min\{|x-n|,|x-(n+1)|\left\}\right)dx=\frac{1}{2}\times 1\times \frac{1}{2}=\frac{1}{4}\text{Now,}{A}_2=\underset{n+1}{\overset{n+2}{\int }}(|x-n|-|x-(n+1\left)|\right)dx=\underset{1}{\overset{2}{\int }}(|t|-|t-1|)dt\text{Put}x=n+t\Rightarrow dx=dt=\underset{1}{\overset{2}{\int }}(t-(t-1\left)\right)dt=\underset{1}{\overset{2}{\int }}1dt=(t{)}_1^2=1\text{and}{A}_3=\underset{n+2}{\overset{n+3}{\int }}(|x-(n+4)|-|x-(n+3\left)|\right)dx=\underset{2}{\overset{3}{\int }}(|t-4|-|t-3|)dt\text{Put}x=n+t\Rightarrow dx=dt=\underset{2}{\overset{3}{\int }}\left(\right(4-t)-(3-t\left)\right)dt=\underset{2}{\overset{3}{\int }}1dt=1\text{Also,}g\left(x\right)=A_1+A_2+A_3=\frac{1}{4}+1+1=\frac{9}{4}\therefore \sum _{n=1}^{100}g\left(x\right)=g\left(1\right)+g\left(2\right)+g\left(3\right)+\cdots +g\left(100\right)=\frac{9}{4}+\frac{9}{4}+\cdots +\frac{9}{4}=\frac{900}{4}$ ​​​​​​