मराठी

For continuity, at x = a, each of aflimx→a+f(x) and aflimx→a-f(x) is equal to f(a). - Mathematics

Advertisements
Advertisements

प्रश्न

For continuity, at x = a, each of `lim_(x -> "a"^+) "f"(x)` and `lim_(x -> "a"^-) "f"(x)` is equal to f(a).

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर

उत्तर

This statement is True.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity And Differentiability - Solved Examples [पृष्ठ १०६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Solved Examples | Q 42 | पृष्ठ १०६

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Examine the following function for continuity:

f (x) = x – 5


Examine the following function for continuity:

`f (x)1/(x - 5), x != 5`


Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`


A function f(x) is defined as,

\[f\left( x \right) = \begin{cases}\frac{x^2 - x - 6}{x - 3}; if & x \neq 3 \\ 5 ; if & x = 3\end{cases}\]  Show that f(x) is continuous that x = 3.

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

(i) \[f\left( x \right) = \begin{cases}\left| x \right| \cos\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{cases}at x = 0\]

 


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

(ii) \[f\left( x \right) = \left\{ \begin{array}{l}x^2 \sin\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]


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

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

Show that 

\[f\left( x \right) = \begin{cases}1 + x^2 , if & 0 \leq x \leq 1 \\ 2 - x , if & x > 1\end{cases}\]


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?

 


Given the function  
\[f\left( x \right) = \frac{1}{x + 2}\] . Find the points of discontinuity of the function f(f(x)).

The function 

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

 


If  \[f\left( x \right) = \begin{cases}\frac{{36}^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, then k equals

 


The function  \[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

 


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  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (ab) for which f (x) cannot be continuous at x = 1, is

 


Let \[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]


The function f (x) = e|x| is


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


If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\]  then f (x) is 


Find the points of discontinuity , if any for the function : f(x) = `(x^2 - 9)/(sinx - 9)`


If f(x) = `(e^(2x) - 1)/(ax)` .                for x < 0 , a ≠ 0
         = 1.                             for x = 0
         = `(log(1 + 7x))/(bx)`.        for x > 0 , b ≠ 0
is continuous at x = 0 . 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`


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 value of k which makes the function defined by f(x) = `{{:(sin  1/x",",  "if"  x ≠ 0),("k"",",  "if"  x = 0):}`, continuous at x = 0 is ______.


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


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


f(x) = `{{:(|x|cos  1/x",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×