हिंदी

Discuss the Continuity of the Function at the Point Given. If the Function is Discontinuous, Then Remove the Discontinuity. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

योग

उत्तर

Given f(0) = 5     ..........(1)
Consider,
`lim_(x ->0) f(x) = lim_(x ->0)(sin^2 5x)/x^2 `

= `lim_(x ->0) ((sin 5x)/(5x) xx 5)^2   (thereforex -> 0, x ≠0)`

= 1 x 5²

= 25

From (i) and (ii) 

`lim_(x ->0) f(x) ≠  f(0)` 

Function is discontinuous at x = 0
Hence we can remove the discontinuity by redefining f as

`f(x) = lim_(x ->0) (sin^2 5x)/x^2` for x ≠ 0

= 25    for x = 0

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2013-2014 (October)

APPEARS IN

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

A function f(x) is defined as 

\[f\left( x \right) = \begin{cases}\frac{x^2 - 9}{x - 3}; if & x \neq 3 \\ 6 ; if & x = 3\end{cases}\]

Show that f(x) is continuous at x = 3

 

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

\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - 1 \right|}{x - 1}, for & x \neq 1 \\ 2 , for & x = 1\end{cases}at x = 1\]

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5


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


If \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if }  x \neq 4 \\ k , & \text{ if }  x = 4\end{cases}\]  is continuous at x = 4, find k.


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 


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 

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

The function f (x) = sin−1 (cos x) is


The function f (x) = e|x| is


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


Let f (x) = |sin x|. Then,


The function f (x) =  |cos x| is


Find the value of k for which the function f (x ) =  \[\binom{\frac{x^2 + 3x - 10}{x - 2}, x \neq 2}{ k , x^2 }\] is continuous at x = 2 .

 
 

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


Prove that the function f defined by 
f(x) = `{{:(x/(|x| + 2x^2)",",  x ≠ 0),("k",  x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.


If f(x) = `{{:("m"x + 1",",  "if"  x ≤ pi/2),(sin x + "n"",",  "If"  x > pi/2):}`, is continuous at x = `pi/2`, then ______.


Write the number of points where f(x) = |x + 2| + |x - 3| is not differentiable.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×