मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Examine the Continuity of the Following Function - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Examine the continuity of the following function :

`{:(,,f(x)= x^2 -x+9,"for",x≤3),(,,=4x+3,"for",x>3):}}"at "x=3`

बेरीज

उत्तर

Given

`f(x)=x^2-x+9 , for x<=3`

` =4x+3        for x>3`

`f(3)=(3)^2-3+9=9-3+9`

`f(3)=15`

`Now lim_(x->3^-)f(x)=lim_(x->3)(x^2-x+9)`

                         ` =(3)^2-(3)+9`

                           =15

`lim_(x->3^-)f(x)=lim_(x->3)(4x+3)`

                        =4(3)+3

                        =15

Thus from the above 

`lim_(x->3^-)f(x)=lim_(x->3)f(x)=15=f(3)`

Hence function is continuous at x=3

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2014-2015 (March)

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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.

For what value of k is the following function continuous at x = 1? \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}, & x \neq 1 \\ k , & x = 1\end{cases}\]


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


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

 


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, & \text{ if } x \neq 0 \\ 10 , & \text{ if }  x = 0\end{cases}\]


Let  \[f\left( x \right) = \begin{cases}\frac{x^4 - 5 x^2 + 4}{\left| \left( x - 1 \right) \left( x - 2 \right) \right|}, & x \neq 1, 2 \\ 6 , & x = 1 \\ 12 , & x = 2\end{cases}\]. Then, f (x) is continuous on the set

 


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

 


Find whether the function is differentiable at x = 1 and x = 2 

\[f\left( x \right) = \begin{cases}x & x \leq 1 \\ \begin{array} 22 - x  \\ - 2 + 3x - x^2\end{array} & \begin{array}11 \leq x \leq 2 \\ x > 2\end{array}\end{cases}\]

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is


If f is continuous at x = 0, then find f (0). 

Where f(x) = `(3^"sin x" - 1)^2/("x" . "log" ("x" + 1)) , "x" ≠ 0`


 If the function f (x) = `(15^x - 3^x - 5^x + 1)/(x tanx)`,  x ≠ 0 is continuous at x = 0 , then find f(0).


Examine the continuity of the following function :
f(x) = x2 - x + 9,          for x ≤ 3
      = 4x + 3,               for x > 3 
at x = 3.


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


Discuss the continuity of the function f at x = 0, where
f(x) = `(5^x + 5^-x - 2)/(cos2x - cos6x),` for x ≠ 0
      = `1/8(log 5)^2,`  for x = 0


Find the value of the constant k so that the function f defined below is continuous at x = 0, where f(x) = `{{:((1 - cos4x)/(8x^2)",", x ≠ 0),("k"",", x = 0):}`


Discuss the continuity of the function f(x) = sin x . cos x.


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


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


The value of k (k < 0) for which the function f defined as

f(x) = `{((1-cos"kx")/("x"sin"x")","  "x" ≠ 0),(1/2","  "x" = 0):}`

is continuous at x = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×