हिंदी

F(x) = ,if,if{|x-4|2(x-4),if x≠40,if x=4 at x = 4 - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग

उत्तर

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

At x = 4

L.H.L. = `lim_(x -> 4^-) |x - 4|/(2(x - 4))`

= `lim_("h" -> 0) (|(4 - "h") - 4|)/(2[(4 - "h") - 4])`

= `lim_("h" -> 0) |-"h"|/(-2"h")`

= `lim_("h" -> 0) |"h"|/(-2"h")`

= `lim_("h" -> 0) "h"/(-2"h")`

= `(-1)/2`

But given that f(4) = 0 ≠ L.H.L.

So, f(x) is discontinuous at x = 4.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ १०७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 5 | पृष्ठ १०७

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

Examine the following function for continuity:

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


Examine the following function for continuity:

f(x) = | x – 5|


If \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\]

Find whether f(x) is continuous at x = 0.

 

Show that 

\[f\left( x \right) = \begin{cases}\frac{\left| x - a \right|}{x - a}, when & x \neq a \\ 1 , when & x = a\end{cases}\] is discontinuous at x = a.

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if }   x \neq 0 \\ 4 , & \text{ if }  x = 0\end{cases}\]

 


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


Find all the points of discontinuity of f defined by f (x) = | x |− | x + 1 |.


Determine if \[f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}\] is a continuous function?

 


Find all point of discontinuity of the function 

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

If f (x) = | x − a | ϕ (x), where ϕ (x) is continuous function, then


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


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

 


The values of the constants ab and c for which the function  \[f\left( x \right) = \begin{cases}\left( 1 + ax \right)^{1/x} , & x < 0 \\ b , & x = 0 \\ \frac{\left( x + c \right)^{1/3} - 1}{\left( x + 1 \right)^{1/2} - 1}, & x > 0\end{cases}\] may be continuous at x = 0, are

 


The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]  


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


Is every continuous function differentiable?


Let \[f\left( x \right) = \begin{cases}\frac{1}{\left| x \right|} & for \left| x \right| \geq 1 \\ a x^2 + b & for \left| x \right| < 1\end{cases}\] If f (x) is continuous and differentiable at any point, then

 

 

 


The function f (x) =  |cos x| is


If f is continuous at x = 0 then find f(0) where f(x) = `[5^x + 5^-x - 2]/x^2`, x ≠ 0


The total cost C for producing x units is Rs (x2 + 60x + 50) and the price is Rs (180 - x) per unit. For how many units the profit is maximum.


Find the value of 'k' if the function 
f(x) = `(tan 7x)/(2x)`,                   for x ≠ 0.
      = k                                        for x = 0.
is continuous at x = 0.


Find `dy/dx if y = tan^-1 ((6x)/[ 1 - 5x^2])`


If the function f is continuous at x = 0 then find f(0),
where f(x) =  `[ cos 3x - cos x ]/x^2`, `x!=0`


If the function f is continuous at x = 2, then find 'k' where

f(x) = `(x^2 + 5)/(x - 1),` for  1< x ≤ 2 
      = kx + 1 , for x > 2


The set of points where the functions f given by f(x) = |x – 3| cosx is differentiable is ______.


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


f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0


f(x) = |x| + |x − 1| 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


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×