English

Prove that the Function F ( X ) = { X | X | + 2 X 2 , X ≠ 0 K , X = 0 Remains Discontinuous at X = 0, Regardless the Choice of K. - Mathematics

Advertisements
Advertisements

Question

Prove that the function 

\[f\left( x \right) = \begin{cases}\frac{x}{\left| x \right| + 2 x^2}, & x \neq 0 \\ k , & x = 0\end{cases}\]  remains discontinuous at x = 0, regardless the choice of k.

Solution

The given function can be rewritten as: 

\[f\left( x \right) = \begin{cases}\frac{x}{x + 2 x^2}, x > 0 \\ \frac{- x}{x - 2 x^2}, x < 0 \\ k, x = 0\end{cases}\]
\[\Rightarrow f\left( x \right) = \begin{cases}\frac{1}{2x + 1}, x > 0 \\ \frac{1}{2x - 1}, x < 0 \\ k, x = 0\end{cases}\]

We observe

(LHL at x = 0) =

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{1}{2x + 1}, x > 0 \\ \frac{1}{2x - 1}, x < 0 \\ k, x = 0\end{cases}\]
\[\lim_{h \to 0} \frac{1}{- 2h - 1} = - 1\]

(RHL at x = 0) =

\[\lim_{h \to 0} \frac{1}{- 2h - 1} = - 1\]
\[\lim_{h \to 0} \frac{1}{2h + 1} = 1\]

So, ​

\[\lim_{h \to 0} \frac{1}{2h + 1} = 1\]

\[\lim_{x \to 0^-} f\left( x \right) \text{and} \lim_{x \to 0^+} f\left( x \right)\]

Thus, f(x) is discontinuous at x = 0, regardless of the choice of k.

 
shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Continuity - Exercise 9.1 [Page 19]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 9 Continuity
Exercise 9.1 | Q 24 | Page 19

RELATED QUESTIONS

Show that the function `f(x)=|x-3|,x in R` is continuous but not differentiable at x = 3.


Is the function f defined by f(x)= `{(x, if x<=1),(5, if x > 1):}`  continuous at x = 0? At x = 1? At x = 2?


Find all points of discontinuity of f, where f is defined by `f(x) = {(|x|+3, if x<= -3),(-2x, if -3 < x < 3),(6x + 2, if x >= 3):}`


Find all points of discontinuity of f, where f is defined by `f (x) = {(x/|x|, if x<0),(-1, if x >= 0):}`


Find all points of discontinuity of f, where f is defined by `f (x) = {(x+1, if x>=1),(x^2+1, if x < 1):}`


Find all points of discontinuity of f, where f is defined by `f (x) = {(x^10 - 1, if x<=1),(x^2, if x > 1):}`


Is the function defined by `f(x) = {(x+5, if x <= 1),(x -5, if x > 1):}` a continuous function?


Show that the function defined by  g(x) = x = [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.


Determine if f defined by `f(x) = {(x^2 sin  1/x, "," if x != 0),(0, "," if x = 0):}` is a continuous function?


Examine the continuity of f, where f is defined by `f(x) = {(sin x - cos x, if x != 0),(-1, "," if x = 0):}`


Find the value of constant ‘k’ so that the function f (x) defined as

f(x) = `{((x^2 -2x-3)/(x+1), x != -1),(k, x != -1):}`

is continous at x = -1


Show that the function f(x) = `{(x^2, x<=1),(1/2, x>1):}` is continuous at x = 1 but not differentiable.


For what value of λ is the function 
\[f\left( x \right) = \begin{cases}\lambda( x^2 - 2x), & \text{ if }  x \leq 0 \\ 4x + 1 , & \text{  if } x > 0\end{cases}\]continuous at x = 0? What about continuity at x = ± 1?


Find the relationship between 'a' and 'b' so that the function 'f' defined by 

\[f\left( x \right) = \begin{cases}ax + 1, & \text{ if }  x \leq 3 \\ bx + 3, & \text{ if } x > 3\end{cases}\] is continuous at x = 3.

 


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

\[f\left( x \right) = \begin{cases}\left| x \right| + 3 , & \text{ if } x \leq - 3 \\ - 2x , & \text { if }  - 3 < x < 3 \\ 6x + 2 , & \text{ if }  x > 3\end{cases}\]

Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}2x , & \text{ if }  & x < 0 \\ 0 , & \text{ if }  & 0 \leq x \leq 1 \\ 4x , & \text{ if }  & x > 1\end{cases}\]


Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}x^{10} - 1, & \text{ if }  x \leq 1 \\ x^2 , & \text{ if } x > 1\end{cases}\]


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if }  & x \geq 1\end{cases}\]


In the following, determine the value of constant involved in the definition so that the given function is continuou: 

\[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x} , & x < \frac{\pi}{2} \\ 3 , & x = \frac{\pi}{2} \\ \frac{3 \tan 2x}{2x - \pi}, & x > \frac{\pi}{2}\end{cases}\]

If f(x) = `{{:("a"x + 1,  "if"  x ≥ 1),(x + 2,  "if"  x < 1):}` is continuous, then a should be equal to ______.


Find all points of discontinuity of the function f(t) = `1/("t"^2 + "t" - 2)`, where t = `1/(x - 1)`


The number of discontinuous functions y(x) on [-2, 2] satisfying x2 + y2 = 4 is ____________.


`lim_("x"-> 0) sqrt(1/2 (1 - "cos"  2"x"))/"x"` is equal to


`f(x) = {{:(x^10 - 1",", if x ≤ 1),(x^2",", if x > 1):}` is discontinuous at


Let a, b ∈ R, b ≠ 0. Define a function

F(x) = `{{:(asin  π/2(x - 1)",", "for"  x ≤ 0),((tan2x - sin2x)/(bx^3)",", "for" x > 0):}`

If f is continuous at x = 0, then 10 – ab is equal to ______.


If f(x) = `{{:(cos ((π(sqrt(1 + x) - 1))/x)/x,",", x ≠ 0),(π/k,",", x = 0):}`

is continuous at x = 0, then k2 is equal to ______.


If f(x) = `{{:((log_(sin|x|) cos^2x)/(log_(sin|3x|) cos  x/2), |x| < π/3; x ≠ 0),(k, x = 0):}`, then value of k for which f(x) is continuous at x = 0 is ______.


If the function f defined as f(x) = `1/x - (k - 1)/(e^(2x) - 1)` x ≠ 0, is continuous at x = 0, then the ordered pair (k, f(0)) is equal to ______.


Find the value(s) of 'λ' if the function

f(x) = `{{:((sin^2 λx)/x^2",", if x ≠ 0  "is continuous at"  x = 0.),(1",", if x = 0):}`


If f(x) = `{{:((kx)/|x|"," if x < 0),(  3","   if x ≥ 0):}` is continuous at x = 0, then the value of k is ______.


The graph of the function f is shown below.

Of the following options, at what values of x is the function f NOT differentiable?


Consider the graph `y = x^(1/3)`


Statement 1: The above graph is continuous at x = 0

Statement 2: The above graph is differentiable at x = 0

Which of the following is correct?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×