मराठी

Examine the following function for continuity: f (x) = x – 5 - Mathematics

Advertisements
Advertisements

प्रश्न

Examine the following function for continuity:

f (x) = x – 5

बेरीज

उत्तर

Let a be a real number, then,

`lim_(x->a^+) f (x) = lim_(h->0) (a + h) - 5 = a - 5`

`lim_(x->a^-) f (x) = lim_(h->0) (a - h) -5 = a - 5`

Also f(a) = a - 5

∵ `lim_(x->a^+) f(x) = lim_(x->a^-) f(x)f(a)`

Hence, the given function f(x) = (x - 5) is continuous.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १५९]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12
पाठ 5 Continuity and Differentiability
Exercise 5.1 | Q 3.1 | पृष्ठ १५९

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Discuss the continuity of the function f, where f is defined by `f(x) = {(2x , ","if x < 0),(0, "," if 0 <= x <= 1),(4x, "," if x > 1):}`


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

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

 


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

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

Show that 

\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{\tan 2x} , if x < 0 \\ \frac{3}{2} , if x = 0 \\ \frac{\log(1 + 3x)}{e^{2x} - 1} , if x > 0\end{cases}\text{is continuous at} x = 0\]


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  \[f\left( x \right) = \begin{cases}kx + 1, if & x \leq 5 \\ 3x - 5, if & x > 5\end{cases}\] at x = 5


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

 


Discuss the continuity of the f(x) at the indicated points:  f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

 

Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.

 

If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\] 

 


Find f (0), so that  \[f\left( x \right) = \frac{x}{1 - \sqrt{1 - x}}\]  becomes continuous at x = 0.

 


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 

The function  \[f\left( x \right) = \frac{x^3 + x^2 - 16x + 20}{x - 2}\] is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as

 


If  \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at  \[x = \frac{\pi}{2}\], if

 


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


If  \[f\left( x \right) = \begin{cases}\frac{\sin \left( \cos x \right) - \cos x}{\left( \pi - 2x \right)^2}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k is equal to


Show that the function 

\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.


Is every differentiable function continuous?


If f (x) is differentiable at x = c, then write the value of 

\[\lim_{x \to c} f \left( x \right)\]

The function f (x) = sin−1 (cos x) is


The function f (x) = e|x| is


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


If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is


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

= 10 for x = 4  at x = 4


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 .

 
 

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.


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 off at x = 1, if

f (x) = 5x - 3 , for 0 ≤ x ≤ 1

       = x2 + 1 , for 1 ≤ x ≤ 2


Examine the continuity of the followin function : 

  `{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "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


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.


Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if"  x ≠ 0),(0",",  "if"  x = 0):}` is discontinuous at x = 0.


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


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


y = |x – 1| is a continuous function.


The set of points where the function f given by f(x) = |2x − 1| sinx is differentiable is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×