English

Find the Points of Discontinuity, If Any, of the Following Functions: F ( X ) = { E X − 1 Log E ( 1 + 2 X ) , If X ≠ 0 7 , If X = 0 - Mathematics

Advertisements
Advertisements

Question

Find the points of discontinuity, if any, of the following functions: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]
Sum
Advertisements

Solution

Given: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]

We have 

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{e^x - 1}{\log_e \left( 1 + 2x \right)} = \lim_{x \to 0} \frac{\left( \frac{e^x - 1}{x} \right)}{\left( \frac{2 \log_e \left( 1 + 2x \right)}{2x} \right)} = \frac{1}{2} \times \frac{\lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)}{\lim_{x \to 0} \left( \frac{\log_e \left( 1 + 2x \right)}{2x} \right)} = \frac{1}{2}\]

It is given that 

\[f\left( 0 \right) = 7\]
⇒  \[\lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)\]

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Continuity - Exercise 9.2 [Page 34]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 9 Continuity
Exercise 9.2 | Q 3.07 | Page 34

RELATED QUESTIONS

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

`f(x)=(15^x-3^x-5^x+1)/(xtanx) , x!=0`


Examine the following function for continuity:

f(x) = `(x^2 - 25)/(x + 5)`, x ≠ −5


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): (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): 

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

If  \[f\left( x \right) = \begin{cases}\frac{1 - \cos kx}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\text{is continuous at x} = 0, \text{ find } k .\]


For what value of k is the function

\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]  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; 

\[f\left( x \right) = \begin{cases}kx + 1, \text{ if }  & x \leq \pi \\ \cos x, \text{ if }  & x > \pi\end{cases}\] at x = π

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

 


Prove that
\[f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}\] is everywhere continuous.

 


Define continuity of a function at a point.

 

If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\]  , then

 


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 

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

 


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


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

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

Discuss the continuity and differentiability of f (x) = e|x| .


Give an example of a function which is continuos but not differentiable at at a point.


Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 


Find k, if f(x) =`log (1+3x)/(5x)` for x ≠ 0

                     = k                    for x = 0

is continuous at x = 0. 


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 .

 
 

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

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


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


If y = ( sin x )x , Find `dy/dx`


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


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):}`


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous at x = 0.


Examine the differentiability of the function f defined by
f(x) = `{{:(2x + 3",",  "if"  -3 ≤ x < - 2),(x + 1",",  "if"  -2 ≤ x < 0),(x + 2",",  "if"  0 ≤ x ≤ 1):}`


The function f(x) = |x| + |x – 1| is ______.


f(x) = `{{:(|x|cos  1/x",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0


f(x) = `{{:((2^(x + 2) - 16)/(4^x - 16)",",  "if"  x ≠ 2),("k"",",  "if"  x = 2):}` at x = 2


Examine the differentiability of f, where f is defined by
f(x) = `{{:(x[x]",",  "if"  0 ≤ x < 2),((x - 1)x",",  "if"  2 ≤ x < 3):}` at x = 2


If f is continuous on its domain D, then |f| is also continuous on D.


`lim_("x" -> "x" //4) ("cos x - sin x")/("x"- "x" /4)`  is equal to ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×