English

A continuous function can have some points where limit does not exist. - Mathematics

Advertisements
Advertisements

Question

A continuous function can have some points where limit does not exist.

Options

  • True

  • False

MCQ
True or False
Advertisements

Solution

This statement is False.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Continuity And Differentiability - Solved Examples [Page 106]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Solved Examples | Q 44 | Page 106

RELATED QUESTIONS

Examine the following function for continuity:

f(x) = `1/(x - 5)`, x ≠ 5


Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{1 - x^n}{1 - x}, & x \neq 1 \\ n - 1 , & x = 1\end{array}n \in N \right.at x = 1\]

Discuss the continuity of the function f(x) at the point x = 0, where  \[f\left( x \right) = \begin{cases}x, x > 0 \\ 1, x = 0 \\ - x, x < 0\end{cases}\]

 


If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\]  is continuous at x = 4, find ab.

 


For what value of k is the function

\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]  continuous at x = 0?

 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  \[f\left( x \right) = \begin{cases}k x^2 , & x \geq 1 \\ 4 , & x < 1\end{cases}\]at x = 1

 


For what value of k is the following function continuous at x = 2? 

\[f\left( x \right) = \begin{cases}2x + 1 ; & \text{ if } x < 2 \\ k ; & x = 2 \\ 3x - 1 ; & x > 2\end{cases}\]

Discuss the continuity of the function  \[f\left( x \right) = \begin{cases}2x - 1 , & \text { if }  x < 2 \\ \frac{3x}{2} , & \text{ if  } x \geq 2\end{cases}\]


Prove that
\[f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}\] is everywhere continuous.

 


Let f (x) = | x | + | x − 1|, then


If  \[f\left( x \right) = \begin{cases}\frac{\sin (a + 1) x + \sin x}{x} , & x < 0 \\ c , & x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{bx\sqrt{x}} , & x > 0\end{cases}\]is continuous at x = 0, then 


The value of f (0), so that the function 

\[f\left( x \right) = \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]   becomes continuous for all x, given by

The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


The value of b for which the function 

\[f\left( x \right) = \begin{cases}5x - 4 , & 0 < x \leq 1 \\ 4 x^2 + 3bx , & 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 

If  \[f\left( x \right) = \begin{cases}a \sin\frac{\pi}{2}\left( x + 1 \right), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\] is continuous at x = 0, then a equals


Show that f(x) = x1/3 is not differentiable at x = 0.


Show that \[f\left( x \right) =\]`{(12x, -,13, if , x≤3),(2x^2, +,5, if x,>3):}` is differentiable at x = 3. Also, find f'(3).


Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.


The set of points where the function f (x) = x |x| is differentiable is 

 


If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text {  is }\] 


If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]


Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4 

= 10 for x = 4  at x = 4


Find k, if the function f is continuous at x = 0, where

`f(x)=[(e^x - 1)(sinx)]/x^2`,      for x ≠ 0

     = k                             ,        for x = 0


If f is continuous at x = 0, then find f (0). 

Where f(x) = `(3^"sin x" - 1)^2/("x" . "log" ("x" + 1)) , "x" ≠ 0`


If the function f is continuous at x = 0

Where f(x) = 2`sqrt(x^3 + 1)` + a,  for x < 0,
= `x^3 + a + b,  for x > 0
and f (1) = 2, then find a and b.


Examine the continuity of the following function :

`{:(,f(x),=(x^2-16)/(x-4),",","for "x!=4),(,,=8,",","for "x=4):}} " at " x=4`


Examine the continuity of the following function :
f(x) = x2 - x + 9,          for x ≤ 3
      = 4x + 3,               for x > 3 
at x = 3.


 If the function f is continuous at x = I, then find f(1), where f(x) = `(x^2 - 3x + 2)/(x - 1),` for x ≠ 1


If the function
f(x) = x2 + ax + b,         x < 2

      = 3x + 2,                 2≤ x ≤ 4

      = 2ax + 5b,             4 < x

is continuous at x = 2 and x = 4, then find the values of a and b


If f(x) = `{{:((x^3 + x^2 - 16x + 20)/(x - 2)^2",", x ≠ 2),("k"",", x = 2):}` is continuous at x = 2, find the value of k.


The number of points at which the function f(x) = `1/(log|x|)` is discontinuous is ______.


f(x) = `{{:(3x + 5",", "if"  x ≥ 2),(x^2",", "if"  x < 2):}` at x = 2


f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2


f(x) = `{{:(|x - 4|/(2(x - 4))",", "if"  x ≠ 4),(0",", "if"  x = 4):}` at x = 4


f(x) = |x| + |x − 1| at x = 1


f(x) = `{{:(3x - 8",",  "if"  x ≤ 5),(2"k"",",  "if"  x > 5):}` at x = 5


Examine the differentiability of f, where f is defined by
f(x) = `{{:(x[x]",",  "if"  0 ≤ x < 2),((x - 1)x",",  "if"  2 ≤ x < 3):}` at x = 2


An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is ______.


If the following function is continuous at x = 2 then the value of k will be ______.

f(x) = `{{:(2x + 1",", if x < 2),(                 k",", if x = 2),(3x - 1",", if x > 2):}`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×