Tag: real numbers

Quizzes (33) Questions (23)

Questions Related to real numbers

Multiple choice absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

If $x$ be real and positive, then the value of
$y = x + \frac{1}{x}$ satisfies

  1. $0 < y \leq 0.5$
  2. $0.5 < y \leq 1$
  3. $1 < y < 2$
  4. $y \geq 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ 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) $

Multiple choice absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

The absolute value of $\dfrac { \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\cos { x+1 }  \right) { e }^{ \sin { x }  }dx }  }{ \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\sin { x-1 }  \right) { e }^{ \cos { x }  }dx }  } $ is equal to 

  1. $e$
  2. $\pi e$
  3. $\dfrac{e}{2}$
  4. $\dfrac{\pi}{e}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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