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

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प्रश्न

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

पर्याय

  • True

  • False

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उत्तर

This statement is False.

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पाठ 5: Continuity And Differentiability - Solved Examples [पृष्ठ १०६]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Solved Examples | Q 44 | पृष्ठ १०६

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Discuss the continuity of the function f, where f is defined by `f(x) = {(3, ","if 0 <= x <= 1),(4, ","if 1 < x < 3),(5, ","if 3 <= x <= 10):}`


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): (iv) \[f\left( x \right) = \left\{ \begin{array}{l}\frac{e^x - 1}{\log(1 + 2x)}, if & x \neq a \\ 7 , if & x = 0\end{array}at x = 0 \right.\]


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

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

In the following, determine the value of constant involved in the definition so that the given function is continuou:   \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if }  0 \leq x \leq 1\end{cases}\]


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


Find all point of discontinuity of the function 

\[f\left( t \right) = \frac{1}{t^2 + t - 2}, \text{ where }  t = \frac{1}{x - 1}\]

Define continuity of a function at a point.

 

\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


If \[f\left( x \right) = \begin{cases}a x^2 - b, & \text { if }\left| x \right| < 1 \\ \frac{1}{\left| x \right|} , & \text { if }\left| x \right| \geq 1\end{cases}\]  is differentiable at x = 1, find a, b.


If f is defined by f (x) = x2, find f'(2).


Discuss the continuity and differentiability of 

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

If \[f\left( x \right) = \left| \log_e |x| \right|\] 


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

= 10 for x = 4  at x = 4


`f(x)=(x^2-9)/(x - 3)` is not defined at x = 3. what value should be assigned to f(3) for continuity of f(x) at = 3?


Discuss the continuity of f at x = 1
Where f(X) = `[ 3 - sqrt ( 2x + 7 ) / ( x - 1 )]`           For x ≠ 1
                    = `-1/3`                                                 For x = 1


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.


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


If f (x) = `(1 - "sin x")/(pi - "2x")^2` , for x ≠ `pi/2` is continuous at x = `pi/4` , then find `"f"(pi/2) .`


Discuss the continuity of the function at the point given. If the function is discontinuous, then remove the discontinuity.

f (x) = `(sin^2 5x)/x^2` for x ≠ 0 
= 5   for x = 0, at x = 0


If Y = tan-1 `[(cos 2x - sin 2x)/(sin2x + cos 2x)]` then find `(dy)/(dx)`


The function f(x) = [x], where [x] denotes the greatest integer function, is continuous at ______.


The function given by f (x) = tanx is discontinuous on the set ______.


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


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


f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1


Examine the differentiability of f, where f is defined by
f(x) = `{{:(1 + x",",  "if"  x ≤ 2),(5 - x",",  "if"  x > 2):}` at x = 2


Given functions `"f"("x") = ("x"^2 - 4)/("x" - 2) "and g"("x") = "x" + 2, "x" le "R"`. Then which of the following is correct?


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