मराठी

Determine the Value of the Constant K So that the Function F ( X ) = { Sin 2 X 5 X , I F X ≠ 0 K , I F X = 0 is Continuous at X = 0 . - Mathematics

Advertisements
Advertisements

प्रश्न

Determine the value of the constant k so that the function

\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0 .\]

 

उत्तर

\[f\left( x \right) = \binom{\frac{\sin2x}{5x}, if x \neq 0}{k, if x = 0}\]

If  

\[f\left( x \right)\]  is continuous at x = 0, then
\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[\lim_{x \to 0} \frac{\sin2x}{5x} = k\]
\[\lim_{x \to 0} \frac{2\sin2x}{5 \times 2x} = k\]
\[\frac{2}{5} \lim_{x \to 0} \frac{\sin2x}{2x} = k\]
\[\frac{2}{5} \times 1 = k\]
\[k = \frac{2}{5}\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Continuity - Exercise 9.1 [पृष्ठ १९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.1 | Q 22 | पृष्ठ १९

व्हिडिओ ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्‍न

Find the relationship between a and b so that the function f defined by `f(x)= {(ax + 1, if x<= 3),(bx + 3, if x  > 3):}` is continuous at x = 3.


Find the values 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` 


Show that the function defined by f (x) = cos (x2) 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 \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]


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.

 


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:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]


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.

 


 then f (x) is continuous for all
\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - x \right|}{x^2 - x}, & x \neq 0, 1 \\ 1 , & x = 0 \\ - 1 , & x = 1\end{cases}\]  then f (x) is continuous for all

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

 


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

Find the values of a and b, if the function f defined by 

\[f\left( x \right) = \begin{cases}x^2 + 3x + a & , & x \leqslant 1 \\ bx + 2 & , & x > 1\end{cases}\] is differentiable at = 1.

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


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


The function f (x) = x − [x], where [⋅] denotes the greatest integer function is


The function f (x) = 1 + |cos x| is


The function \[f\left( x \right) = \frac{\sin \left( \pi\left[ x - \pi \right] \right)}{4 + \left[ x \right]^2}\] , where [⋅] denotes the greatest integer function, is


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


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


If `f`: R → {0, 1} is a continuous surjection map then `f^(-1) (0) ∩ f^(-1) (1)` is:


A real value of x satisfies `((3 - 4ix)/(3 + 4ix))` = α – iβ (α, β ∈ R), if α2 + β2 is equal to


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) = 5x – 3 is continuous at x =


For what value of `k` the following function is continuous at the indicated point

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×