Tag: continuity

Quizzes (2) Questions (26)

Questions Related to continuity

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

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. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function f(x) = [2x^3 - 5] is discontinuous where the expression inside the bracket is an integer. For x in [1, 2), 2x^3 - 5 ranges from 2(1)^3 - 5 = -3 to 2(2)^3 - 5 = 11. The integer values are -3, -2, ..., 10. There are 14 such values.

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

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

  1. $x=0$ only
  2. $x=2$ only
  3. $x=0$ and $2$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The limit exists as 1 if |x-1| > 1 (i.e., x > 2 or x < 0) and as -1 if |x-1| < 1 (i.e., 0 < x < 2). At x=0 and x=2, the limit does not exist because the left and right limits differ, causing discontinuity.

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

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. ${e}^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To make the function continuous at x=1, the limit as x approaches 1 of the given expression must exist and equal f(1). Evaluating the limit using logarithmic properties and standard forms leads to an evaluation of e^2. Therefore, f(1) must be defined as e^2 to remove the discontinuity.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The greatest integer function [x] is discontinuous at integers, but at those same integer points, the cosine term evaluates to zero because cos(pi(2n-1)/2) = cos(n*pi - pi/2) = 0. The product of the integer value and zero yields zero continuously across all points. Thus, the function is continuous everywhere.

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

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. $\dfrac {23\pi}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

f(x) = [3 sin x] is discontinuous when 3 sin x is an integer. In [0, 2pi], 3 sin x takes integer values -3, -2, -1, 0, 1, 2, 3. Solving 3 sin x = k for k in {-2, -1, 0, 1, 2} yields multiple points. Summing these points leads to 13pi.

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

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 $x=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The limit of (sin x - x cos x) / x^2 as x approaches 0 is 0. Since the function is defined as 0 at x=0, and the limit exists and equals the function value, the discontinuity is removable.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $x=0$.

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

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. $\dfrac{3\pi}{4},\dfrac{7\pi}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The inner function g(x) = sin x + cos x changes values around the points where sin x + cos x = 0, which occurs at 3pi/4 and 7pi/4 in the interval (0, 2pi). Since f(u) has a jump discontinuity at u = 0, the composite function (fog)(x) will be discontinuous where g(x) = 0. Solving g(x) = 0 yields 3pi/4 and 7pi/4.

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

$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 $0,1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The function f(x) = min(x, x^2) compares the line y = x and the parabola y = x^2. For x < 0, x^2 > x, so f(x) = x. For 0 <= x <= 1, x^2 <= x, so f(x) = x^2. For x > 1, x^2 > x, so f(x) = x. Checking the intersection points at x = 0 and x = 1, both pieces meet continuously, making the function continuous on all of R.