Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

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

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

The value of x on simplifying $x -2 |x|=-3$ is

  1. -1 or 3

  2. 1 or -3

  3. -1 or -3

  4. 1, 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x-2|x| = -3$

There two cases possible either $x$ is positive or$x$ is negative
We have to solve both cases
Firstly taking x as positive
$x-2x=-3$
$x=3$
Now taking $x$ as negative
$x-2(-x)=-3$
$3x=-3$
$x=-1$
So we get two values of $x$ that is -1 and 3
 So correct answer will be option A

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

If $a$ and $b$ are any real numbers, then which of the following expressions is always positive?

  1. $\left| a \right| $
  2. $\left| a+b \right| $
  3. $\left| a-b \right| +1/2$
  4. ${ a }^{ 2 }+{ b }^{ 2 }$
  5. ${ \left( a+b \right) }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Lets check each option one by one.

A. |a| cab be 0 if a=0 
B. |a+b| can be 0 if a+b=0
C. |a-b|+ 1/2  will always positive because |a+b| is either 0 or positive 
D. ${a}^{2}+{b}^{2}$ can be 0 if both $a$ and $b$ becomes 0
E ${(a+b)}^{2}$ can equal to 0 if $a+b$ = 0
So correct answer will be option C