Question

If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $a\le t\le x$, $a\le x\le b$}=f(x) & min {f(t), $a\le t\le x$, $a\le x\le b$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $a\le t\le x$, $a\le x\le b$}=f(a),
min {f(t), $a\le t\le x$, $a\le x\le b$}=f(x).
On the basis of above information answer the following questions.

The value of ${\int }_{-100}^{100}min\{\left\{x\right\},\{-x\}\}dx$ equals, where {x} denotes fractional part of x

• A. 200
• B. 100
• C. 50
• D. 0

Answer 1

Answer: (C)

50

${\int }_{-100}^{100}min\{\left\{x\right\},\{-x\}\}dx$
$=2{\int }_{-100}^{100}{\int }_0^{100}min\{\left\{x\right\},\{-x\}\}dx$
$=200{\int }_0^{1}min\{\left\{x\right\},\{-x\}\}dx$
$=200\times \{\begin{array}{cc}x& 0\le x\le \frac{1}{2}\\ -x+1& \frac{1}{2}\le x\le 1\end{array}$
$=200\times$area of $△AOB$
$=200\times \frac{1}{2}\times 1\times \frac{1}{2}$