Tag: mathematics and statistics

Questions Related to mathematics and statistics

If $f(x)=\dfrac {1}{x^{2}-17x+66}$ then $f\left(\dfrac {2}{x-2}\right)$ is discontinuous at $x=$

  1. $2,\dfrac {7}{3},\dfrac {25}{11}$

  2. $2,\dfrac {7}{3},\dfrac {24}{11}$

  3. $2,\dfrac {8}{3},\dfrac {24}{11}$

  4. $None\ of\ these$


Correct Option: A

The sum of all values of $x$ for which $f(x)=[3\sin x]$ is discontinous in $[0,\ 2\pi]$ is (where [.] represents greatest integers function)

  1. $\dfrac {21\pi}{2}$

  2. $13\ \pi$

  3. $11\ \pi$

  4. $\dfrac {23\pi}{2}$


Correct Option: B

Consider the function defined on $[0,\ 1]\rightarrow R,\ f(x)=\dfrac {\sin x-x\cos x}{x^{2}}$ if $x\neq 0$ and $f(0)=0$ then the function of $f(x)$. 

  1. Has a removable discontinuity at $x=0$

  2. Has a removable finite discontinuity at $x=0$

  3. Has a non removable infinite discontinuity at $x=0$

  4. Is continuous at $x=0$


Correct Option: A

The function $f(x)={ sin }^{ -1 }(cosx)$ is :

  1. Discontinuous at x = 0

  2. Continuous at x = 0

  3. Differentiable at x = 0

  4. None of these


Correct Option: A
Explanation:

$f(x)=\sin^{-1} (\cos x)$

LHL :
$lim _{x\rightarrow 0^-} \sin^{-1}(\cos x)=lim _{h\rightarrow 0} \sin^{-1}(\cos (0-h))$

$lim _{h\rightarrow 0} \sin^{-1} (\cos (-h))=lim _{h\rightarrow 0} \sin^{-1}(\cos 0)=\dfrac{\pi}{2}$

RHL:
$lim _{x\rightarrow 0^{+}} \sin^{-1}(\cos x)$
$=lim _{h\rightarrow 0} \sin^{-1} \cos (0+h)$
$=\sin^{-1} \cos 0$
$=\dfrac{\pi}{2}$

Thus, $LHL=RHL=f(0)=\dfrac{\pi}{2}$

RHD :
$lim _{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{sin^{-1}(\cos h)-1}{h}$

$lim _{h\rightarrow 0} \dfrac{\sin^{-1}(\cos h)-1}{h}=\dfrac{1-\sin h}{\sqrt{1-\cos^2 h}}=\dfrac{-\sin h}{\sin h}=-1$

LHD :
$lim _{h\rightarrow 0} \dfrac{f(x-h)-f(x)}{-h}=lim _{h\rightarrow 0}\dfrac{\sin^{-1}(\cos -h)-1}{-h}$

$lim _{h\rightarrow 0} \dfrac{\sin^{-1}(\cos h)-1}{-h}=\dfrac{-\sin h}{-\sin h}=1$

$LHD \neq RHD$
Thus, function is not differentiable at $x=0$.

If $f\left( x \right) =\begin{cases} -1,if\ x<0\ \ 0,if\ x=0\ \ 1,if\ x>0\ \end{cases}$ and $g\left(x\right)=\sin x +\cos x$, then point discontinuity of $(fog)(x)$ in $(0,2\pi)$ are 

  1. $\dfrac{\pi}{4},\dfrac{5\pi}{4}$

  2. $\dfrac{\pi}{4},\dfrac{3\pi}{4}$

  3. $\dfrac{\pi}{4},\dfrac{7\pi}{4}$

  4. $\dfrac{3\pi}{4},\dfrac{7\pi}{4}$


Correct Option: A

$f(x)=\min { \left{ x,{ x }^{ 2 } \right} ,\forall x\epsilon R } $ then $f(x)$ is 

  1. discontinuous at $0$

  2. discontinuous at $1$

  3. continuous on $R$

  4. continuous on $0,1$


Correct Option: A

If $f(x)=\dfrac{1}{1-x}$, the number of points of discontinuity of $f\left{f[f(x)]\right}$ is:

  1. $2$

  2. $1$

  3. $0$

  4. $infinite$


Correct Option: A

Consider  $f ( x ) = \sin x \forall x \in \left[ 0 , \dfrac { \pi } { 2 } \right] ; f ( x ) + f ( \pi - x ) =2 \forall x \in \left( \dfrac { \pi } { 2 } , \pi \right) \text { and } f ( x ) = f ( 2 \pi - x ) \forall x \in ( \pi , 2 \pi ) . \text { If } n , m$ denotes number of points where  $f(x)$  is discontinuous and non derivable respectively in  $[ 0,2 \pi ]$  then value of  $n \div  m$  is

  1. $0$

  2. $1$

  3. $2$

  4. $4$


Correct Option: A

f(x) = $\dfrac{\sin2x + 1}{\sin x - \cos x}$ is discontinuous at $x =$ ____________.

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

  2. $\dfrac{\pi}{3}$

  3. $\dfrac{\pi}{6}$

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


Correct Option: A
Explanation:

Given the function  $f(x)=\dfrac{\sin2x + 1}{\sin x - \cos x}$.

The function will be undefined when,
$\sin x-\cos x=0$
or, $\tan x=1$
or, $x=n\pi+\dfrac{\pi}{4}$. [ $n=0,\pm 1,\pm 2,.....$].
Corresponding to  $n=0$ we get $x=\dfrac{\pi}{4}$.
So this is the point where the function is not defined and hence is not continuous.

The function $\displaystyle f\left ( x \right )=\frac{\log \left ( 1+ax \right )-\log \left ( 1-bx \right )}{x}$ is not defined at $ x = 0$. The value which should be assigned to $f$ at $x =0$ so that it is continuous there, is

  1. $a-b$

  2. $a+b$

  3. $\log a+ \log b$

  4. none of these


Correct Option: B
Explanation:

$f\left( x \right) =\dfrac { \log  \left( 1+ax \right) -\log  \left( 1-bx \right)  }{ x } \ \lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1+ax \right) -\log  \left( 1-bx \right)  }{ x } =\lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1+ax \right)  }{ ax\times \cfrac { 1 }{ a }  } +\lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1bx \right)  }{ -bx\times \cfrac { 1 }{ b }  } \ =a+b$