मराठी

F(x) = k,if,if{1-coskxxsinx, if x≠012, if x=0 at x = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

f(x) = `{{:((1 - cos "k"x)/(xsinx)",",   "if"  x ≠ 0),(1/2",",  "if"  x = 0):}` at x = 0

बेरीज

उत्तर

We have f(x) = `{{:((1 - cos "k"x)/(xsinx)",",   "if"  x ≠ 0),(1/2",",  "if"  x = 0):}` 

Since, f(x) is continuous at x = 0

∴ f(0) = `lim_(x -> 0) "f"(x)`

∴ `1/2 = lim_(x -> 0) (1 - cos "k"x)/(xsinx)`

= `lim_(x -> 0) (1 - cos^2"k"x)/(xsinx) * 1/(1 + cos "k"x)`

= `lim_(x -> 0) (sin^2"k"x)/(xsinx) * 1/(1 + cos"k"x)`

= `lim_(x -> 0) (((sin "k"x)/("k"x))^2 "k"^2)/((sinx)/x) * 1/(1 + cos "k"x)`

= `"k"^2/1 * 1/(1 + 1)`

= `"k"^2/2`

⇒ k2 – 1

⇒ k = ±1

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ १०८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 14 | पृष्ठ १०८

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Examine the following function for continuity:

f (x) = x – 5


A function f(x) is defined as,

\[f\left( x \right) = \begin{cases}\frac{x^2 - x - 6}{x - 3}; if & x \neq 3 \\ 5 ; if & x = 3\end{cases}\]  Show that f(x) is continuous that x = 3.

If \[f\left( x \right) = \begin{cases}e^{1/x} , if & x \neq 0 \\ 1 , if & x = 0\end{cases}\] find whether f is continuous at x = 0.


Discuss the continuity of the function f(x) at the point x = 1/2, where \[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\] 


Determine the values of a, b, c for which the function f(x) = `{((sin(a + 1)x + sin x)/x, "for"   x < 0),(x, "for"  x = 0),((sqrt(x + bx^2) - sqrtx)/(bx^(3"/"2)), "for"  x > 0):}` is continuous at x = 0.


For what value of k is the function

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

 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  \[f\left( x \right) = \begin{cases}kx + 1, if & x \leq 5 \\ 3x - 5, if & x > 5\end{cases}\] at x = 5


For what value of k is the following function continuous at x = 2? 

\[f\left( x \right) = \begin{cases}2x + 1 ; & \text{ if } x < 2 \\ k ; & x = 2 \\ 3x - 1 ; & x > 2\end{cases}\]

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if }  x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if }  x < 1\end{cases}\]


Find all the points of discontinuity of f defined by f (x) = | x |− | x + 1 |.


Find all point of discontinuity of the function 

\[f\left( t \right) = \frac{1}{t^2 + t - 2}, \text{ where }  t = \frac{1}{x - 1}\]

The points of discontinuity of the function 

\[f\left( x \right) = \begin{cases}2\sqrt{x} , & 0 \leq x \leq 1 \\ 4 - 2x , & 1 < x < \frac{5}{2} \\ 2x - 7 , & \frac{5}{2} \leq x \leq 4\end{cases}\text{ is } \left( \text{ are }\right)\] 


Show that f(x) = |x − 2| is continuous but not differentiable at x = 2. 


Give an example of a function which is continuos but not differentiable at at a point.


Write the points where f (x) = |loge x| is not differentiable.


If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text {  is }\] 


If \[f\left( x \right) = \left| \log_e |x| \right|\] 


Find the points of discontinuity , if any for the function : f(x) = `(x^2 - 9)/(sinx - 9)`


Examine the continuity off at x = 1, if

f (x) = 5x - 3 , for 0 ≤ x ≤ 1

       = x2 + 1 , for 1 ≤ x ≤ 2


If y = ( sin x )x , Find `dy/dx`


If f (x) = `(1 - "sin x")/(pi - "2x")^2` , for x ≠ `pi/2` is continuous at x = `pi/4` , then find `"f"(pi/2) .`


If the function
f(x) = x2 + ax + b,         x < 2

      = 3x + 2,                 2≤ x ≤ 4

      = 2ax + 5b,             4 < x

is continuous at x = 2 and x = 4, then find the values of a and b


Find the value of the constant k so that the function f defined below is continuous at x = 0, where f(x) = `{{:((1 - cos4x)/(8x^2)",", x ≠ 0),("k"",", x = 0):}`


Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if"  x ≠ 0),(0",",  "if"  x = 0):}` is discontinuous at x = 0.


A continuous function can have some points where limit does not exist.


f(x) = `{{:(3x + 5",", "if"  x ≥ 2),(x^2",", "if"  x < 2):}` at x = 2


f(x) = `{{:(|x - 4|/(2(x - 4))",", "if"  x ≠ 4),(0",", "if"  x = 4):}` at x = 4


A function f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f′(0) = 2. Prove that f′(x) = 2f(x).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×