рд╣рд┐рдВрджреА

Find the value of k so that the function f is continuous at the indicated point. f(x) = {ЁЭСШтБвЁЭСе + 1, if ЁЭСе тЙд ЁЭЬЛ, cos тБбЁЭСе, if ЁЭСе > ЁЭЬЛ, at x = ╧А

Advertisements
Advertisements

рдкреНрд░рд╢реНрди

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

f(x) = `{(kx +1", if"  x<= pi),(cos x", if"  x > pi):}` at x = π

рдпреЛрдЧ
Advertisements

рдЙрддреНрддрд░

f(x) = `{(kx +1", if"  x<= pi),(cos x", if"  x > pi):}`

If f(x) is continuous at x = π it implies:

f(π) = `lim_(x -> pi^+)` f(x) = `lim_(x -> pi^-)` f(x)

⇒ k(π) + 1 = cos(π)

⇒ k(π) + 1 = −1

⇒ k(π) = −2

⇒ k = `(-2)/pi`

That is, for the quantity k = `(-2)/pi` this function is continuous at x = π.

shaalaa.com
  рдХреНрдпрд╛ рдЗрд╕ рдкреНрд░рд╢реНрди рдпрд╛ рдЙрддреНрддрд░ рдореЗрдВ рдХреЛрдИ рддреНрд░реБрдЯрд┐ рд╣реИ?
рдЕрдзреНрдпрд╛рдп 5: Continuity and Differentiability - Exercise 5.1 [рдкреГрд╖реНрда резремрез]

APPEARS IN

рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдЕрдзреНрдпрд╛рдп 5 Continuity and Differentiability
Exercise 5.1 | Q 28 | рдкреГрд╖реНрда резремрез

рд╡реАрдбрд┐рдпреЛ рдЯреНрдпреВрдЯреЛрд░рд┐рдпрд▓VIEW ALL [3]

рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНрди

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.


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.


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


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.

 


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.

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:  \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if }  x \neq 0 \\ 3k , & \text{ if  } x = 0\end{cases}\] 


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


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


If  \[f\left( x \right) = \binom{\frac{1 - \cos x}{x^2}, x \neq 0}{k, x = 0}\]  is continuous at x = 0, find k


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.


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  . 

 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 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 function 

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

 


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

 


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

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

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


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


If f.g is continuous at x = a, then f and g are separately continuous at x = a.


`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 ____________.


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


The function f(x) = 5x – 3 is continuous at x =


The function f(x) = x2 – sin x + 5 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


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

Englishрд╣рд┐рдВрджреАрдорд░рд╛рдареА


      Forgot password?
Use app×