English

If F ( X ) = { 2 X 2 + K , If X ≥ 0 − 2 X 2 + K , If X < 0 Then What Should Be the Value of K So that F(X) is Continuous at X = 0.

Advertisements
Advertisements

Question

If  \[f\left( x \right) = \begin{cases}2 x^2 + k, &\text{ if }  x \geq 0 \\ - 2 x^2 + k, & \text{ if }  x < 0\end{cases}\]  then what should be the value of k so that f(x) is continuous at x = 0.

 

Sum
Advertisements

Solution

The given function can be rewritten as 

\[f\left( x \right) = \begin{cases}2 x^2 + k, &\text{ if }  x \geq 0 \\ - 2 x^2 + k, & \text{ if }  x < 0\end{cases}\] 

We have
(LHL at x = 0) = 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} - 2 \left( - h \right)^2 + k = k\]

(RHL at x = 0) =\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( 2 h^2 + k \right) = k\]

If

\[f\left( x \right)\] is continuous at
\[x = 0\]
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = k\]

∴ can be any real number.

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Continuity - Exercise 9.1 [Page 21]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 8 Continuity
Exercise 9.1 | Q 41 | Page 21

RELATED QUESTIONS

For what value of λ is the function defined by f(x) = `{(λ(x^2 - 2x)", if"  x <= 0),(4x+ 1", if"  x > 0):}` continuous at x = 0? What about continuity at x = 1?


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

f(x) = `{(kx^2", if"  x<= 2),(3", if"  x > 2):}` at x = 2


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.


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.


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


Let  \[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}\] x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

 


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.


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.

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.

 


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

\[f\left( x \right) = \begin{cases}x^3 - x^2 + 2x - 2, & \text{ if }x \neq 1 \\ 4 , & \text{ if } x = 1\end{cases}\]

 


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


In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if  }  x \leq 2 \\ x - 1, & \text{ if }  x > 2\end{cases}\]


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


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.


If \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}\]

for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].


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

\[\lim_{x \to a}\] f (x) = f (a)?

If \[f\left( x \right) = \begin{cases}\frac{\sin^{- 1} x}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, write the value of k.


If \[f\left( x \right) = \begin{cases}\frac{1 - \sin x}{\left( \pi - 2x \right)^2} . \frac{\log \sin x}{\log\left( 1 + \pi^2 - 4\pi x + 4 x^2 \right)}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k =

 


If  \[f\left( x \right) = \begin{cases}\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] and f (x) is continuous at x = 0, then the value of k is


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 value of f (0), so that the function

\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by 


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  \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]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 


The function f (x) = |cos x| is


Let f (x) = |cos x|. Then,


The function f (x) = 1 + |cos x| is


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


`lim_("x"->0) (1 - "cos x")/"x"`  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


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 ______.


Discuss the continuity of the following function:

f(x) = sin x – cos x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×