Tag: mathematics and statistics

Questions Related to mathematics and statistics

The function 
  $\displaystyle f\left ( x \right )=\frac{\cos x-\sin x}{\cos 2x}$  is not defined at $\displaystyle x=\frac{\pi }{4}$. The value of $\displaystyle  f\left ( \frac{\pi }{4}\right )$ so that $ f\left ( x \right)$  is continuous everywhere, is

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. 1

  2. -1

  3. $\sqrt{2}$

  4. $\displaystyle \frac{1}{\sqrt{2}}$

Show answer
Correct Option: D
Explanation:

The function will be continuous at $x=\dfrac { \pi  }{ 4 } $ if $f\left( \dfrac { \pi  }{ 4 }\right) =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ f\left( x \right) } $.

$f\left( \dfrac { \pi  }{ 4 }  \right) =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ f\left( x \right)  } =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { \cos { x } -\sin { x }  }{ \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x }  }  } \ =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { \cos { x } -\sin { x }  }{ \left( \cos { x } -\sin { x }  \right) \left( \cos { x } +\sin { x }  \right)  }  } $


For $x\neq \frac { \pi  }{ 4 } ,\quad \cos { x } -\sin { x } \neq 0$, so the limit becomes

$\ =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { 1 }{ \cos { x } +\sin { x }  }  } =\frac { 1 }{ \cos { \frac { \pi  }{ 4 }  } +\sin { \frac { \pi  }{ 4 }  }  } =\frac { 1 }{ \sqrt { 2 }  } $

The continuity on an interval has a geometric interpretation. namely, a function f defined on an interval I is continuous on I if its graph has no 'holes' or 'jumps' .f is said to have a removable discontinuity at c if f(x) has a limit at c but lim $\lim _{x\rightarrow c}f\left ( x \right )\neq f\left ( c \right )$. 

If $\lim _{x\rightarrow c+}f\left ( x \right ) and \lim _{x\rightarrow c-}f\left ( x \right )$ exist but are not equal then c is called jump discontinuity. 
If $\lim _{x\rightarrow c+}f\left ( x \right ) and \lim _{x\rightarrow c-}f\left ( x \right )$ fail to exist then c is called infinite discontinuity.

Let $\displaystyle g\left ( x \right )=\begin{cases}x^{2}+5 & \, x< 2 \ 10 & \, x=2 \ 1+x^{3} & \, x> 2 \end{cases}$ then $x=2$ :

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. a point of continuity

  2. is a removable discontinuity

  3. is a jump discontinuity

  4. is.an infinite discontinuity

Show answer
Correct Option: B
Explanation:

$\displaystyle \lim _{ x\rightarrow 2+ }{ g\left( x \right) =1+{ 2 }^{ 3 } } =9$

$\displaystyle \lim _{ x\rightarrow 2- }{ g\left( x \right)  } =4+5=9$

So $\lim _{ x\rightarrow 2 }{ g\left( x \right)  } $ exists but is not equal to $g\left( 2 \right) =10$ 

Thus at $x=2$ is removable discontinuity.

$\displaystyle g(x)= \begin{cases}1 & \, x\leq -2 \ \displaystyle \frac{1}{2} x& \, -2< x< 4 \ \sqrt{x} & , x\geq 4 \end{cases}$.then

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $g$ is a continuous function

  2. all the discontinuities are removable discontinuities

  3. all the discontinuities are jump

  4. all the discontinuities are infinite

Show answer
Correct Option: C
Explanation:

$\displaystyle \lim _{ x\rightarrow 4+ }{ g\left( x \right) } =\displaystyle \lim _{ x\rightarrow 4- }{ g\left( x \right) } =2$


so $f$ is continuous at $x=4$

But $\displaystyle \lim _{ x\rightarrow 2- }{ g\left( x \right)  } =1$ and $\lim _{ x\rightarrow 2+ }{ g\left( x \right)  } =-1$

So $x=-2$ is a jump discontinuity and there is no other discontinuities.

Given $\displaystyle f(x) = \begin{cases} 3-\left [ \cot ^{-1}\left ( \frac{2x^{3}-3}{x^{2}} \right ) \right ] & \mbox{for } x> 0 \ \left { x^{2} \right }\cos \left ( e^{1/x} \right ) & \mbox{for } x< 0 \end{cases}$ where { } & [ ] denotes the fractional part and the integral part functions respectively, then which of the following statement does not hold good -

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $f(0^-)=0$

  2. $f(0^+)=3$

  3. $f(0)=0\Rightarrow :continuity:of:f:at:x=0$

  4. irremovable discontinuity of f at $x=0$

Show answer
Correct Option: A,B,D
Explanation:

RHL $=\displaystyle \lim _{x\rightarrow0^+}(3-[\cot^{-1}(\frac{2{x}^{3}-3}{x^2})])$
$=3-[\cot^{-1}(-\infty)]=3-0=3$
LHL $=\displaystyle \lim _{h\rightarrow0}\left { (0-h)^{2} \right }\cos \left (e^{(\dfrac{1}{0-h})}  \right )$
$=\displaystyle \lim _{h\rightarrow0}(0-h)^2\cos(e^{-\infty})=0$
Clearly discontinuity is irremovable, (since L.H.L$\neq $ R.H.L)

Consider $\displaystyle  f(x) = \begin{cases} x\left [ x \right ]^{2}\log _{(1+x)}2& \mbox{ for } -1< x< 0 \ \dfrac{\ln e^{x^{2}}+2\sqrt{\left { x \right }}}{\tan \sqrt{x}} & \mbox{ for } 0< x< 1  \end{cases}$ where [] & {} are the greatest integer function & fractional part function respectively, then -

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $f(0)=ln2\Rightarrow :f:is:continuous:at:x=0$

  2. $f(0)=2\Rightarrow :f:is:continuous:at:x=0$

  3. $f(0)=e^2\Rightarrow :f:is:continuous:at:x=0$

  4. f has an irremovable discontinuity at $x=0$

Show answer
Correct Option: D
Explanation:

LHL 

$=\displaystyle \lim _{h\rightarrow 0}(0-h)\left [ 0-h \right ]^{2}\log _{(1+0-h)}2$

$=\displaystyle \lim _{h\rightarrow 0}\frac{-h(-1)^{2}\ln 2}{\ln (1-h)}=\ln 2$

RHL

$=\displaystyle \lim _{h\rightarrow 0}\frac{\ln \left ( e^{(0+h)^{2}}

\right )+2\sqrt{\left { 0+h \right }}}{\tan \sqrt{(0+h)}}$


$=\displaystyle

\lim _{h\rightarrow 0}\frac{\ln \left ( e^{h^{2}}+2\sqrt{h} \right

)}{\tan \sqrt{(h)}}=\lim _{h\rightarrow 0}\frac{\ln \left (

e^{h^{2}}+2\sqrt{h} \right )}{\sqrt{(h)}}$

$=\displaystyle

\lim _{h\rightarrow 0}\left [ h^{2}+\dfrac{\ln \left (

1+\dfrac{2\sqrt{h}}{e^{h^{2}}} \right

).\dfrac{2\sqrt{h}}{e^{h^{2}}}}{\dfrac{2\sqrt{h}}{e^{h^{2}}}} \right

]\dfrac{1}{\sqrt{h}}=2$

$\therefore  RHL\neq LHL\Rightarrow $ f has an irremovable discontinuity.

Let $f(x)=\begin{cases} \dfrac{1-\cos 2x}{2x^2}&:& x\ne 0\k &:& x=0 \end{cases}$.

Then the value of $k$ for which, $f(x)$ will be continuous at $x=0$ is

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $0$

  2. $1$

  3. $2$

  4. none of these

Show answer
Correct Option: B
Explanation:

If $f(x)$ is to be continuous at $x=0$ then we must have, $\lim\limits _{x\to 0}f(x)=f(0)$.


$f(0)=k$

Now,

$\lim\limits _{x\to 0}f{(x)}\\$
$=\lim\limits _{x\to 0}\dfrac{1-\cos 2x}{2x^2}\\$
$=\lim\limits _{x\to 0}\dfrac{2\sin^2 x}{2x^2}\\$
$=\left(\lim\limits _{x\to 0}\dfrac{\sin x}{x}\right)^2\\$
$=1$.

So if $f(x)$ to be continuous at $x=0$ then $k$ should be $1$.

Let $f\left( x \right) =\dfrac { \log { \left( 1+x+{ x }^{ 2 } \right)  } +\log { \left( 1-x+{ x }^{ 2 } \right)  }  }{ \sec { x } -\cos { x }  } ,x\neq 0$ The value of $f\left (0\right)$ so that $f$ is continuous at $x=0$ is 

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $1$

  2. $0$

  3. $2$

  4. None of these

Show answer
Correct Option: A

Given $f(x)=\dfrac{\left[ \left{ \left| x \right|  \right}  \right] { e }^{ { x }^{ 2 } }\left{ \left[ \left| x+\left{ x \right}  \right|  \right]  \right} }{\left( { e }^{ 1/{ x }^{ 2 } }-1 \right) sgn\left( \sin { x }  \right) }$ for $x\neq 0$
$=0, for\  x=0$
Where $\left{ x \right} $ is the fractional part function; $[x]$ is the step up function and $sgn{(x)}$ is the signum function of $x$ then, $f(x)$

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. is continuous at $x=0$

  2. is discontinuous at $x=1$

  3. has a removable discontinuity at $x=0$

  4. has an irremovable discontinuity at $x=0$

Show answer
Correct Option: A
Explanation:
$\dfrac { \left[ \left\{ \left| x \right|  \right\}  \right] { e }^{ { x }^{ 2 } }\left\{ \left[ \left| x+\left\{  \right\}  \right|  \right]  \right\}  }{ \left( { e }^{ \dfrac { 1 }{ { x }^{ 2 } }  }-1 \right) sgn\left( \sin { x }  \right)  } $
$f(o)=\dfrac{continous}{continuous}=continuous$
$f(o)=continuous$
$f(x)$ is continuous at $x=0$

Given that $\displaystyle \prod _{n=1}^n cos \dfrac{x}{2^n}= \dfrac{\sin  x}{2^n  \sin \left ( \dfrac{x}{2^n} \right )}$ and $\displaystyle f(x) = \left{\begin{matrix}\lim _{n \rightarrow \infty}\sum _{n = 1}^n \dfrac{1}{2^n} \tan \left (\dfrac{x}{2^n} \right ), & x \in (0, \pi) - \left {\dfrac{\pi}{2} \right }\ \dfrac{2}{\pi} & x = \dfrac{\pi}{2}\end{matrix}\right.$
Then which one of the following is true?

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $f(x)$ has non-removable discontinuity of finite type at $\displaystyle x = \dfrac{\pi}{2}$.

  2. $f(x)$ has removable discontinuity at $\displaystyle x = \dfrac{\pi}{2}$

  3. $f(x)$ is continuous at $\displaystyle x = \dfrac{\pi}{2}$.

  4. $f(x)$ has non-removable discontinuity of infinite type at $\displaystyle x = \dfrac{\pi}{2}$

Show answer
Correct Option: C
Explanation:
We know that $\tan { x } =\cfrac { 2\tan { \left( \cfrac { x }{ 2 }  \right)  }  }{ 1-\tan ^{ 2 }{ \left( \cfrac { x }{ 2 }  \right)  }  } $
$\Rightarrow \cot { x } =\cfrac { 1-\tan ^{ 2 }{ \left( \cfrac { x }{ 2 }  \right)  }  }{ 2\tan { \left( \cfrac { x }{ 2 }  \right)  }  } =\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cfrac { 1 }{ 2 } \tan { \left( \cfrac { x }{ 2 }  \right)  } $
$\Rightarrow \cfrac { 1 }{ 2 } \tan { \left( \cfrac { x }{ 2 }  \right)  } =\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cot { x } \rightarrow 1$
Replacing x by $\cfrac { x }{ 2 } $ and multiplying by $\cfrac { 1 }{ 2 } $
$\cfrac { 1 }{ 4 } \tan { \left( \cfrac { x }{ 4 }  \right)  } =\cfrac { 1 }{ 4 } \cot { \left( \cfrac { x }{ 4 }  \right)  } -\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } $
Repeating the steps
$\cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } =\cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cfrac { 1 }{ { 2 }^{ n-1 } } \cot { \left( \cfrac { x }{ { 2 }^{ n-1 } }  \right)  } $
$\therefore \sum _{ n=1 }^{ n }{ \left[ \cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  \right]  } =\left[ \cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cot { x }  \right] +\left[ \cfrac { 1 }{ 4 } \cot { \left( \cfrac { x }{ 4 }  \right)  } -\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  }  \right] +....+\left[ \cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cfrac { 1 }{ { 2 }^{ n-1 } } \cot { \left( \cfrac { x }{ { 2 }^{ n-1 } }  \right)  }  \right] $
$=\cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cot { x } =\cfrac { \left( \cfrac { 1 }{ { 2 }^{ n } }  \right)  }{ \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  } -\cot { x } $
$f\left( x \right) =\lim _{ n\rightarrow \infty  }{ \sum _{ n=1 }^{ n }{ \left[ \cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  \right]  }  } =\lim _{ n\rightarrow \infty  }{ \left( \cfrac { \left( \cfrac { x }{ { 2 }^{ n } }  \right) \times \cfrac { 1 }{ x }  }{ \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  } -\cot { x }  \right)  } $
as $n\rightarrow \infty ,\cfrac { 1 }{ { 2 }^{ n } } \rightarrow 0$
$\cfrac { x }{ { 2 }^{ n } } \rightarrow 0$
$f\left( x \right) =\cfrac { 1 }{ x } \times 1-\cot { x } $
$x=\cfrac { \pi  }{ 2 } $
$f\left( \cfrac { \pi  }{ 2 }  \right) =\cfrac { 2 }{ \pi  } $ (given )$\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ f\left( x \right)  } =\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ \left( \cfrac { 1 }{ x } -\cot { x }  \right)  } $
$=\cfrac { 1 }{ \cfrac { \pi  }{ 2 }  } -0=\cfrac { 2 }{ \pi  } $
$\therefore f\left( \cfrac { \pi  }{ 2 }  \right) =\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ f\left( x \right)  } $
$\Rightarrow f\left( x \right) $ is continuous at $x=\cfrac { \pi  }{ 2 }$

The value of f(0) so that the function
$f(x)=\displaystyle \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$
becomes continuous, is equal to

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $\dfrac{1}{6}$

  2. $\dfrac{1}{4}$

  3. 2

  4. $\dfrac{1}{3}$

Show answer
Correct Option: A
Explanation:

The function $f(x)$ is continuous except possibly at $x=0$. For $f$ to be continuous at $x=0$, we must have
$\displaystyle f(0)=\lim _{x\rightarrow 0}f(x)=\lim _{x\rightarrow 0}\frac{(1+x)^{1/2}-(1+x)^{1/3}}{x}$

$=\displaystyle \lim _{x\rightarrow 0}\dfrac{\left [ 1+\dfrac{1}{2}x+\dfrac{(1/2)(-1/2)}{2}x^{2}+... \right ]-\left [ 1+\dfrac{1}{3}x+\dfrac{(1/3)(-2/3)}{2}x^{2}+... \right ]}{x}$
Therefore,
$\displaystyle f(0)=\lim _{x\rightarrow 0}x\dfrac{\left [ \dfrac{1}{2}+\dfrac{(1/2)(-1/2)}{2}x+... \right ]-\left [\dfrac{1}{3}+\dfrac{(1/3)(-2/3)}{2}x+... \right ]}{x}$


$=\displaystyle \lim _{x\rightarrow 0}\left [ \left ( \dfrac{1}{2}-\dfrac{1}{3} \right )+ term\  containing\  x \right ]=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$
(Alternatively apply L-Hospital's Rule)