Tag: real numbers (rational and irrational numbers)

$|-10| = 10$ and $|28| = 28$

$|0(-9)| = |0| = 0$

Rule: $|x|=x$ when $x$ is positive integer and $|x|=-x$ when $x$ is negative integer. So

$|7|+|5|-|7|-|1-3|$

$ Given\quad y\quad =\quad x+\frac { 1 }{ x } \ \quad =(\sqrt { x } )^{ 2 }+\left( \frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\ \quad =\left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\sqrt { x } \frac { 1 }{ \sqrt { x }  } \ \quad =\left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\ First\quad term\quad is\quad a\quad squared\quad term\quad so\quad it\quad is\quad positive.\ \therefore \quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\quad >2 \quad when\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\quad has\quad a\quad finite\quad value\ and\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\quad =\quad 2\quad \quad  when\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\quad is\quad zero.\ \therefore \quad y\ge 2\quad \quad (Ans) $

If the numbers are consecutive, they can be represented as x, x+1, and x+2. Since x+2 is the largest, x is the smallest.

$\begin{array}{l} We\, have \ I=\dfrac { { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \sin  x } }\left( { x\cos  x+1 } \right) dx }  } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { x\sin  x-1 } \right) dx }  } }  \ =\dfrac { { \left[ { x{ e^{ \sin  x } } } \right] _{ 0 }^{ \frac { \pi  }{ 2 }  } } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { 1-x\sin  x } \right) dx }  } } =\dfrac { { \frac { \pi  }{ 2 } \times e } }{ { \left[ { { e^{ \cos  x } }x } \right] _{ \frac { \pi  }{ 2 }  }^{ 0 } } }  \ =\dfrac { { \frac { \pi  }{ 2 } e } }{ { 0-\frac { \pi  }{ 2 }  } } =-e \ Hence,\, absolute\, value\, =e \ Hence,\, option\, A\; is\, the\, correct\, answer. \end{array}$

Absolute value is always a positive value.