Removable Discontinuities

Q. Rolle's theorem is not applicable in which one of the following cases?

1. $f\left(x\right)=|x|\text{in}[-2,2]$
2. $f\left(x\right)=x^2-4x+5\text{in}[1,3]$
3. $f\left(x\right)=\left[x\right]\text{in}[2.5,2.7]$
4. $f\left(x\right)=x^2-x\text{in}[0,1]$

Q. Which of the following functions $f$ has/ have a removable discontinuity at the indicated point?

1. $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
2. $f\left(x\right)=\frac{x-7}{|x-7|}$ at $x=7$
3. $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
4. $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$

Q. Let $f\left(x\right)=a{x}^2+bx+c,$ where $a$ is positive and $b$ and $c$ are both negative, then the number of point in $R$ where $g\left(x\right)=f\left(|x|\right)$ is non-differentiable is/are

Q. Let $f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x& 2x\\ \sin x& x& x\end{array}\right|$, then $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}$ is equal to

(a) 0
(b) −1
(c) 2
(d) 3

Q. If $y=\left|\begin{array}{ccc}f\left(x\right)& g\left(x\right)& h\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$ prove that

$\frac{dy}{dx}=\left|\begin{array}{ccc}{f}^{\prime }\left(x\right)& g^{\prime }\left(x\right)& h^{\prime }\left(x\right)\\ l& m& n\\ a& b& c\end{array}\right|$

Q. Let $f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x^2},\mathrm{if}x\ne 0\\ 1,\mathrm{if}x=0\end{array}$. Which of the following is/are true?

1. f(x) will be continuous at x = 0 if f(x) = $\frac{1}{2}$ at x = 0
   
2. f(x) has a removable discontinuity.
3. $\underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}.$
   
4. f(x) has a non-removable discontinuity

Q. If $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin (a+1)x+\sin x}{x},& x<0\\ c,& x=0\\ \frac{\sqrt{x+b{x}^2}-\sqrt{x}}{bx\sqrt{x}},& x>0\end{array}\right.$is continuous at x = 0, then
(a) a = $-\frac{3}{2}$, b = 0, c = $\frac{1}{2}$
(b) a = $-\frac{3}{2}$, b = 1, c = $-\frac{1}{2}$
(c) a = $-\frac{3}{2}$, b ∈ R − {0}, c = $\frac{1}{2}$
(d) none of these

Q.

You can remove a removable discontinuity by

1. By redefining the domain of function appropriately

2. By redefining the function value at the point of discontinuity

3. By excluding the points where function goes to infinity

4. All of these

Q. Let $f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x^2},\mathrm{if}x\ne 0\\ 1,\mathrm{if}x=0\end{array}$. Which of the following is/are true?

1. f(x) will be continuous at x = 0 if f(x) = $\frac{1}{2}$ at x = 0
   
2. f(x) has a removable discontinuity.
3. $\underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}.$
   
4. f(x) has a non-removable discontinuity

Q. $\underset{0}{\overset{1}{\int }}\frac{2x}{1+x^4}dx$

Q. Q46


$$\begin{gathered} \mathbf{46}\mathbf{.}Ify=\frac{\sqrt{1-{\sin }^{-1}x}}{1+{\sin }^{-1}x}then{y}^1\left(0\right)= \\ a)0b)1 \\ c)-1d)noneofthese \end{gathered}$$

Q.

In the plot of the function below. Which is the point at which the discontinuity is of removable type?

1. q

2. r

3. All the above

4. p

Q. A discontinuous function can be made continuous by redefining the domain to exclude the points of discontinuities.

1. False

2. True

Q.

You can remove a removable discontinuity by

1. By redefining the domain of function appropriately

2. By redefining the function value at the point of discontinuity

3. By excluding the points where function goes to infinity

4. All of these

Q. If $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\cos kx}{x\sin x},& x\ne 0\\ \frac{1}{2},& x=0\end{array}\right.\mathrm{is}\mathrm{continuous}\mathrm{at}x=0,\mathrm{find}k.$

Q. If $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin \left(\cos x\right)-\cos x}{{\left(\mathrm{\pi }-2x\right)}^2},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right.$is continuous at x = π/2, then k is equal to
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) −1

Q. Which of the following functions $f$ has/ have a removable discontinuity at the indicated point?

1. $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
2. $f\left(x\right)=\frac{x-7}{|x-7|}$ at $x=7$
3. $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
4. $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$

Q.

In the plot of the function below. Which is the point at which the discontinuity is of removable type?

1. p

2. q

3. r

4. All the above

Q. The function $f\left(x\right)=|x|+\frac{|x|}{x}$ is

1. Discontinuous at origin because $|x|$ is discontinuous there
2. Discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there.
3. Continuous at origin
4. Discontinuous at origin because both $|x|$ and $\frac{|x|}{x}$ are discontinuous there

Q. Show that $f\left(x\right)=\left\{\begin{array}{c}\frac{\sin 3x}{\tan 2x},\mathrm{if}x<0\\ \frac{3}{2},\mathrm{if}x=0\\ \frac{\log (1+3x)}{e^{2x}-1},\mathrm{if}x>0\end{array}\right.\mathrm{is}\mathrm{continuous}\mathrm{at}x=0$

Q. Which of the following functions $f$ has/ have a removable discontinuity at the indicated point?

1. $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
2. $f\left(x\right)=\frac{x-7}{|x-7|}$ at $x=7$
3. $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
4. $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$