English

Show that the Function G (X) = X − [X] is Discontinuous at All Integral Points. Here [X] Denotes the Greatest Integer Function.

Advertisements
Advertisements

Question

Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Sum
Advertisements

Solution

Given: `g(x)=x-[x]`

It is evident that g is defined at all integral points.

Let \[n \in Z\] .

Then, 

`g(n)=n-[n]=n-n=0`

The left hand limit of at x = n is,

`lim_(x->n^-)g(x)=lim_(x->n^_)(x-[x])=lim_(x->n^_)(x) -lim_(x->n^_)[x]=n-(n-1)=1`

The right hand limit of f at n is, 

`lim_(x->n^+)g(x)=lim_(x->n^+)(x-[x])=lim_(x->n^+)(x)-lim_(x->n^+)[x]=n-n=0`

It is observed that the left and right hand limits of f at x = n do not coincide.

i.e.

\[\lim_{x \to n^-} g\left( x \right) \neq \lim_{x \to n^+} g\left( x \right)\]

So, f is not continuous at x = n,

\[n \in Z\]

Hence, g is discontinuous at all integral points.

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Continuity - Exercise 9.2 [Page 37]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 8 Continuity
Exercise 9.2 | Q 12 | Page 37

RELATED QUESTIONS

A function f (x) is defined as
f (x) = x + a, x < 0
= x,       0 ≤x ≤ 1
= b- x,   x ≥1
is continuous in its domain.
Find a + b.


If f (x) is continuous on [–4, 2] defined as 

f (x) = 6b – 3ax, for -4 ≤ x < –2
       = 4x + 1,    for –2 ≤ x ≤ 2

Show that a + b =`-7/6`


Is the function defined by f(x) = x2 − sin x + 5 continuous at x = π?


Find the value of k so that the function f is continuous at the indicated point.

f(x) = `{((kcosx)/(pi-2x)", if"  x != pi/2),(3", if"  x = pi/2):}` at x = `"pi/2`


Find the value of k so that the function f is continuous at the indicated point.

f(x) = `{(kx + 1", if"  x <= 5),(3x - 5", if"  x > 5):}` at x = 5


Show that the function defined by f(x) = |cos x| is a continuous function.


Find the values of a so that the function 

\[f\left( x \right) = \begin{cases}ax + 5, if & x \leq 2 \\ x - 1 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]

Extend the definition of the following by continuity 

\[f\left( x \right) = \frac{1 - \cos7 (x - \pi)}{5 (x - \pi )^2}\]  at the point x = π.

If  \[f\left( x \right) = \frac{2x + 3\ \text{ sin }x}{3x + 2\ \text{ sin }  x}, x \neq 0\] If f(x) is continuous at x = 0, then find f (0).


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if }  & x = 0\end{cases}\] at x = 0


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

 


In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]


In the following, determine the value of constant involved in the definition so that the given function is continuou:   \[f\left( x \right) = \begin{cases}4 , & \text{ if } x \leq - 1 \\ a x^2 + b, & \text{ if }  - 1 < x < 0 \\ \cos x, &\text{ if }x \geq 0\end{cases}\]


The function  \[f\left( x \right) = \begin{cases}x^2 /a , & \text{ if } 0 \leq x < 1 \\ a , & \text{ if } 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \text{ if }  \sqrt{2} \leq x < \infty\end{cases}\] is continuous on (0, ∞), then find the most suitable values of a and b.


Show that f (x) = cos x2 is a continuous function.


What happens to a function f (x) at x = a, if  

\[\lim_{x \to a}\] f (x) = f (a)?

If \[f\left( x \right) = \begin{cases}\frac{x}{\sin 3x}, & x \neq 0 \\ k , & x = 0\end{cases}\]  is continuous at x = 0, then write the value of k.


Determine whether \[f\left( x \right) = \binom{\frac{\sin x^2}{x}, x \neq 0}{0, x = 0}\]  is continuous at x = 0 or not.

 


Determine the value of the constant 'k' so that function 

\[\left( x \right) = \begin{cases}\frac{kx}{\left| x \right|}, &\text{ if }  x < 0 \\ 3 , & \text{ if } x \geq 0\end{cases}\]  is continuous at x  = 0  . 

Let  \[f\left( x \right) = \left\{ \begin{array}\\ \frac{x - 4}{\left| x - 4 \right|} + a, & x < 4 \\ a + b , & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, & x > 4\end{array} . \right.\]Then, f (x) is continuous at x = 4 when

 

 


The function  \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\] 


The function 

\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & x \neq 0 \\ \frac{k}{2} , & x = 0\end{cases}\]  is continuous at x = 0, then k =

Let  \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}, x \neq \frac{\pi}{4} .\]  The value which should be assigned to f (x) at  \[x = \frac{\pi}{4},\]so that it is continuous everywhere is


If the function f (x) defined by  \[f\left( x \right) = \begin{cases}\frac{\log \left( 1 + 3x \right) - \log \left( 1 - 2x \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then k =

 


If  \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]then the value of the function at = 0, so that the function is continuous at x = 0, is

 


The value of a for which the function \[f\left( x \right) = \begin{cases}5x - 4 , & \text{ if } 0 < x \leq 1 \\ 4 x^2 + 3ax, & \text{ if } 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 


Find the values of a and b so that the function

\[f\left( x \right)\begin{cases}x^2 + 3x + a, & \text { if } x \leq 1 \\ bx + 2 , &\text {  if } x > 1\end{cases}\] is differentiable at each x ∈ R.

If is defined by  \[f\left( x \right) = x^2 - 4x + 7\] , show that \[f'\left( 5 \right) = 2f'\left( \frac{7}{2} \right)\] 


Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if


`lim_("x" -> 0) (1 - "cos" 4 "x")/"x"^2` is equal to ____________.


Let `"f" ("x") = ("In" (1 + "ax") - "In" (1 - "bx"))/"x", "x" ne 0` If f (x) is continuous at x = 0, then f(0) = ____________.


Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0


The function f(x) = x2 – sin x + 5 is continuous at x =


Find the values of `a` and ` b` such that the function by:

`f(x) = {{:(5",", if  x ≤ 2),(ax + b",", if 2 < x < 10),(21",", if x ≥ 10):}`

is a continuous function.


The value of ‘k’ for which the function f(x) = `{{:((1 - cos4x)/(8x^2)",",  if x ≠ 0),(k",",  if x = 0):}` is continuous at x = 0 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×