हिंदी

Examine the Continuity of the Function F ( X ) = { 3 X − 2 , X ≤ 0 X + 1 , X > 0 a T X = 0 Also Sketch the Graph of this Function. - Mathematics

Advertisements
Advertisements

प्रश्न

Examine the continuity of the function  

\[f\left( x \right) = \left\{ \begin{array}{l}3x - 2, & x \leq 0 \\ x + 1 , & x > 0\end{array}at x = 0 \right.\]

Also sketch the graph of this function.

योग
Advertisements

उत्तर

Meaning of continuity of function - If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x-coordinate of the point at which continuity is to be checked

If: 

`lim_(h->0)f(c - h) = lim_(h->0)f(c + h) = f(c)`

where h is a very small positive no (can assume h = 0.00000000001 like this)

It means :- 

Limiting value of the left neighbourhood of x = c also called left hand limit LHL `{i.e lim_(h->0)f(c - h)}` must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL `{i.e lim_(h->0)f(c + h)}` and both must be equal to the value of f(x) at x = c i.e. f(c). 

Thus, it is the necessary condition for a function to be continuous.

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Lets solve now:

Given function is

f(x) = `{(3x - 2, if x ≤ 0),(x + 1, if x > 0):}`         ...(2)

We need to check whether f(x) is continuous at x=0 or not

For this we need to check LHL, RHL and value of function at x = 0

Clearly,

f(0) = 3*0 - 2 = -2 [from equation 2] 

LHL = `lim_(h->0)f(0 - h) = lim_(h->0)f(-h) = lim_(h->0){3(-h)-2} = -2`

RHL = `lim_(h->0)f(0 + h) = lim_(h->0)f(h) = lim_(h->0){h + 1} = 0 + 1 = 1`

As, LHL ≠ RHL

f(x) is discontinuous at x = 0

This can also be proved by plotting f(x) on cartesian plane.

For x >0, we need to plot

y = x + 1

put y = 0, we get x = -1 and for second point we put x = 0 and thus get y=1

two points are enough to plot the straight line.

Two coordinates are (-1,0) and (0,1)

For x ≤ 0, we need to plot 

y = 3x - 2

put x = 0 then y = -2

on putting y = 0 we get x 2/3

two coordinates are (0, -2) and `(2/3, 0)`

Graph:

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Continuity - Exercise 9.1 [पृष्ठ १८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.1 | Q 14 | पृष्ठ १८

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Discuss the continuity of the following function:

f(x) = sin x × cos x


Discuss the continuity of the cosine, cosecant, secant and cotangent functions.


Find the values of a and b such that the function defined by f(x) = `{(5", if"  x <= 2),(ax +b", if"  2 < x < 10),(21", if"  x >= 10):}` is a continuous function.


Find the values of a so that the function 

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

Find the value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]


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


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


Discuss the continuity of the function  

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

Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if }  x = 2\end{cases}\]


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if }  x < 0 \\ 2x + 3, & x \geq 0\end{cases}\]


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if }  x = 0\end{cases}\]


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.


The function f(x) is defined as follows: 

\[f\left( x \right) = \begin{cases}x^2 + ax + b , & 0 \leq x < 2 \\ 3x + 2 , & 2 \leq x \leq 4 \\ 2ax + 5b , & 4 < x \leq 8\end{cases}\]

If f is continuous on [0, 8], find the values of a and b.


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


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.


If the function   \[f\left( x \right) = \frac{\sin 10x}{x}, x \neq 0\] is continuous at x = 0, find f (0).

 


If  \[f\left( x \right) = \frac{1 - \sin x}{\left( \pi - 2x \right)^2},\] when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x= π/2, where λ =


The value of a for which the function \[f\left( x \right) = \begin{cases}\frac{\left( 4^x - 1 \right)^3}{\sin\left( x/a \right) \log \left\{ \left( 1 + x^2 /3 \right) \right\}}, & x \neq 0 \\ 12 \left( \log 4 \right)^3 , & x = 0\end{cases}\]may be continuous at x = 0 is

 


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 \[f\left( x \right) = \begin{cases}\frac{1 - \cos 10x}{x^2} , & x < 0 \\ a , & x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + \sqrt{x}} - 25}, & x > 0\end{cases}\] then the value of a so that f (x) may be continuous at x = 0, is 


If  \[f \left( x \right) = \sqrt{x^2 + 9}\] , write the value of

\[\lim_{x \to 4} \frac{f\left( x \right) - f\left( 4 \right)}{x - 4} .\]

If \[f\left( x \right) = \begin{cases}\frac{\left| x + 2 \right|}{\tan^{- 1} \left( x + 2 \right)} & , x \neq - 2 \\ 2 & , x = - 2\end{cases}\]  then f (x) is


Let f (x) = |cos x|. Then,


The function f (x) = 1 + |cos x| is


If f(x) = 2x and g(x) = `x^2/2 + 1`, then which of the following can be a discontinuous function ______.


The function f(x) = `(4 - x^2)/(4x - x^3)` is ______.


Let f(x) = |sin x|. Then ______.


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

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


`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) = ____________.


If `f(x) = {{:(-x^2",", "when"  x ≤ 0),(5x - 4",", "when"  0 < x ≤ 1),(4x^2 - 3x",", "when"  1 < x < 2),(3x + 4",", "when"  x ≥ 2):}`, then


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


What is the values of' 'k' so that the function 'f' is continuous at the indicated point


Discuss the continuity of the following function:

f(x) = sin x – cos x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×