Tag: mathematics and statistics

Questions Related to mathematics and statistics

Find the slope and $y$-intercept of the line $x - y = 3$

  1. slope $= 2$, $y$-intercept $= -3$

  2. slope $= 0$, $y$-intercept $= -3$

  3. slope $= 1$, $y$-intercept $= -3$

  4. slope $= 1$, $y$-intercept $= 3$


Correct Option: C
Explanation:
The slope intercept form of the line is $y=mx+c$ where $m$ is the slope of the line and $c$ is the $y$-intercept.
Change the equation $x-y=3$ in slope intercept form:
$x-y=3$
$\Rightarrow -y=-x+3$
$\Rightarrow y=x-3$
Hence, the slope of the line $x-y=3$ is $m=1$ and the $y$-intercept is $-3$.

A line in the $xy$-plane passes through the origin and has a slope of $\dfrac{1}{7}$. Which of the following points lies on the line?

  1. $\left(0, 7\right)$

  2. $\left(1, 7\right)$

  3. $\left(7, 7\right)$

  4. $\left(14, 2\right)$


Correct Option: D
Explanation:

If any straight line passes through origin, then it must of the form $y = mx$.


Now if the slope is $\dfrac{1}{7}$, then line will be $y = \dfrac{1}{7}x \ $ or $ \ 7y -x = 0$

We can see out of all the points only point $(14,2)$ satisfies the equation of the line. Hence Only $(14,2)$ lies on the line.

Correct option is $D$

Find an equation of the line through the points $(-3,5)$ and $(9,10)$ and write it in standard form $Ax+By=C$, with $A>0$

  1. $6x-10y=-75$

  2. $5x-12y=-75$

  3. $4x-11y=-65$

  4. $x-6y=-15$


Correct Option: B
Explanation:
Given points are $(-3,5)$ and $(9,10)$
The slope of the line is given by:
$m =\dfrac{ (10-5)}{[9-(-3)] }= \dfrac {5}{12}$
The equation becomes: 
$y - 5 = \left (\dfrac {5}{12}\right) [x-(-3)]$ 
$y-5=\left (\dfrac {5}{12}\right)(x+3)$ 
Solve it and get the equation- 
$12y-60=5x+15$ 
$5x-12y=-75$ 

Convert $40^\circ \,20'$ into radian measure.

  1. $\dfrac {121}{540}\pi $ radians

  2. $\dfrac {121}{570}\pi $ radians

  3. $\dfrac {120}{513}\pi $ radians

  4. None


Correct Option: A
Explanation:
Given: ${40^0}$${{{20}^{'}}}$

${40^0} + \dfrac{{{{20}^0}}}{{{{60}^0}}}  $

$=40 + \dfrac{1^o}{3} = \dfrac{{{{121}^0}}}{3}$

$radian = \dfrac{\pi }{{180^o}} \times \dfrac{{121^o}}{3}$

$\boxed{ = \dfrac{{121}}{{540}}\pi \;radians}$

Find the radian measure corresponding to the degree $-47^{o}30'$

  1. $\dfrac {-19\ \pi}{72}rad$

  2. $\dfrac {19\ \pi}{72}rad$

  3. $\dfrac {13\ \pi}{72}rad$

  4. $None\ of\ these$


Correct Option: A
Explanation:
$-47^{o} 30'$
$\Rightarrow - (47+ \dfrac{30}{60}) (\because 1^{o} =60')$
$\Rightarrow  -\left( 47+ \dfrac{1}{2} \right)$
$-\left( \dfrac{95}{2} \right)$
Radian measure $\Rightarrow \dfrac{\pi}{180} \times \dfrac{-95}{2}$
$\Rightarrow \pi x - \dfrac{19}{72} \Rightarrow - \dfrac{19 \pi}{72}$ radian

The value of $\tan\left(7\dfrac{1}{2}\right)^o$ is 

  1. $\dfrac {2\sqrt 2 -(1+\sqrt 3)}{\sqrt 3-1}$

  2. $\dfrac {1+\sqrt 3}{1-\sqrt 3}$

  3. $\dfrac {1}{\sqrt 3}+\sqrt 3$

  4. $\sqrt 2 +\sqrt 3$


Correct Option: A
Explanation:

$\tan\left(7\dfrac{1}{2}\right)^o=\tan\left(\dfrac{15}{2}\right)^o$


                       $=\dfrac{\sin\dfrac{15^o}{2}}{\cos\dfrac{15^o}{2}}$

                       $=\dfrac{2\sin\dfrac{15^o}{2}\times \sin\dfrac{15^o}{2}}{2\sin\dfrac{15^o}{2}\times \cos\dfrac{15^o}{2}}$

                        $=\dfrac{2\sin^2\dfrac{15^o}{2}}{\sin\left(2\times\dfrac{15^o}{2}\right)}$

                        $=\dfrac{1-\cos\left(2\times\dfrac{15^o}{2}\right)}{\sin 15^o}$

                        $=\dfrac{1-\cos 15^o}{\sin 15^o}$

Now,
$\cos 15^o=\cos(45^o-30^o)$
              $=\cos 45^o\cos30^o+\sin 45^o\sin 30^o$
              $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
              $=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\sin 15^o=\sin(45^o-30^o)$
             $=\sin 45^o\cos30^o-\cos 45^o\sin 30^o$
             $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
             $=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$


$\tan\left(7\dfrac{1}{2}\right)^o=\dfrac{1-\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{\dfrac{\sqrt{3}-1}{2\sqrt{2}}}$

                      $=\dfrac{2\sqrt{2}-(\sqrt{3}+1)}{\sqrt{3}-1}$

Number of points of discontinuity of $f\left( x \right) = \left[ {2{x^3} - 5} \right]$ in $\left[ {1,2} \right)$ is where $\left[ x \right]$ denotes greatest integer function are

  1. $14$

  2. $13$

  3. $10$

  4. $8$


Correct Option: A

$f(x)=\displaystyle\lim _{n\rightarrow \infty}\dfrac{(x-1)^{2n}-1}{(x-1)^{2n}+1}$ is discontinuous at

  1. $x=0$ only

  2. $x=2$ only

  3. $x=0$ and $2$

  4. None of these


Correct Option: C

If $f\left( x \right) ={ \left( \tan { \left( \dfrac { \pi  }{ 4 } +\ell nx \right)  }  \right)  }^{ \log _{ x }{ e }  }$ is to be made continuous at $X=1$, then $f(1)$ should be equal to

  1. ${e}^{2}$

  2. $e$

  3. $1/e$

  4. ${e}^{-2}$


Correct Option: B

The function $f\left( x \right)=\left[ x \right] \cos { \left( \pi \left( \dfrac { 2x-1 }{ 2 }  \right)  \right)  } $. (where [.] denotes the greatest integer function ) is discontinuous.  

  1. For each real $x$

  2. For each integral point

  3. No where

  4. At each non-integral point


Correct Option: A