मराठी

Write an Example of a Function Which is Everywhere Continuous but Fails to Differentiable Exactly at Five Points. - Mathematics

Advertisements
Advertisements

प्रश्न

Write an example of a function which is everywhere continuous but fails to differentiable exactly at five points.

थोडक्यात उत्तर
Advertisements

उत्तर

\[f\left( x \right) = \left| x \right| + \left| x + 1 \right| + \left| x + 2 \right| + \left| x + 3 \right| + \left| x + 4 \right|\]

The above function is continuous everywhere but not differentiable at x = 0, 1, 2, 3 and 4

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Differentiability - Exercise 10.2 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 10 Differentiability
Exercise 10.2 | Q 8 | पृष्ठ १६

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

If f(x)= `{((sin(a+1)x+2sinx)/x,x<0),(2,x=0),((sqrt(1+bx)-1)/x,x>0):}`

is continuous at x = 0, then find the values of a and b.


Examine the following function for continuity:

f(x) = `1/(x - 5)`, x ≠ 5


Examine the following function for continuity:

f(x) = `(x^2 - 25)/(x + 5)`, 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.

A function f(x) is defined as 

\[f\left( x \right) = \begin{cases}\frac{x^2 - 9}{x - 3}; if & x \neq 3 \\ 6 ; if & x = 3\end{cases}\]

Show that f(x) is continuous at x = 3

 

If \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}; for & x \neq 1 \\ 2 ; for & x = 1\end{cases}\] Find whether f(x) is continuous at x = 1.

 


Show that 

\[f\left( x \right) = \begin{cases}\frac{\left| x - a \right|}{x - a}, when & x \neq a \\ 1 , when & x = a\end{cases}\] is discontinuous at x = a.

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}(x - a)\sin\left( \frac{1}{x - a} \right), & x \neq a \\ 0 , & x = a\end{array}at x = a \right.\]

 


Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{2\left| x \right| + x^2}{x}, & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]

Find the value of k for which \[f\left( x \right) = \begin{cases}\frac{1 - \cos 4x}{8 x^2}, \text{ when}  & x \neq 0 \\ k ,\text{ when }  & x = 0\end{cases}\] is continuous at x = 0;

 


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: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]

Find the values of a and b so that the function f(x) defined by \[f\left( x \right) = \begin{cases}x + a\sqrt{2}\sin x , & \text{ if }0 \leq x < \pi/4 \\ 2x \cot x + b , & \text{ if } \pi/4 \leq x < \pi/2 \\ a \cos 2x - b \sin x, & \text{ if }  \pi/2 \leq x \leq \pi\end{cases}\]becomes continuous on [0, π].


Given the function  
\[f\left( x \right) = \frac{1}{x + 2}\] . Find the points of discontinuity of the function f(f(x)).

The function 

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

 


If f (x) = | x − a | ϕ (x), where ϕ (x) is continuous function, then


The value of f (0), so that the function 

\[f\left( x \right) = \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]   becomes continuous for all x, given by

Show that \[f\left( x \right) =\]`{(12x, -,13, if , x≤3),(2x^2, +,5, if x,>3):}` is differentiable at x = 3. Also, find f'(3).


Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: 

\[f\left( x \right) = \begin{cases}3x - 2, & 0 < x \leq 1 \\ 2 x^2 - x, & 1 < x \leq 2 \\ 5x - 4, & x > 2\end{cases}\]

Find whether the function is differentiable at x = 1 and x = 2 

\[f\left( x \right) = \begin{cases}x & x \leq 1 \\ \begin{array} 22 - x  \\ - 2 + 3x - x^2\end{array} & \begin{array}11 \leq x \leq 2 \\ x > 2\end{array}\end{cases}\]

Let f (x) = |sin x|. Then,


Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 


Examine the continuity of f(x)=`x^2-x+9  "for"  x<=3`

=`4x+3  "for"  x>3,  "at"  x=3` 


Find k, if f(x) =`log (1+3x)/(5x)` for x ≠ 0

                     = k                    for x = 0

is continuous at x = 0. 


Evaluate :`int Sinx/(sqrt(cos^2 x-2 cos x-3)) dx`


The probability distribution function of continuous random variable X is given by
f( x ) = `x/4`,  0 < x < 2
        = 0,       Otherwise
Find P( x ≤ 1)


Let f(x) = `{{:((1 - cos 4x)/x^2",",  "if"  x < 0),("a"",",  "if"  x = 0),(sqrt(x)/(sqrt(16) + sqrt(x) - 4)",", "if"  x > 0):}`. For what value of a, f is continuous at x = 0?


Examine the differentiability of the function f defined by
f(x) = `{{:(2x + 3",",  "if"  -3 ≤ x < - 2),(x + 1",",  "if"  -2 ≤ x < 0),(x + 2",",  "if"  0 ≤ x ≤ 1):}`


The function f(x) = |x| + |x – 1| is ______.


The value of k which makes the function defined by f(x) = `{{:(sin  1/x",",  "if"  x ≠ 0),("k"",",  "if"  x = 0):}`, continuous at x = 0 is ______.


For continuity, at x = a, each of `lim_(x -> "a"^+) "f"(x)` and `lim_(x -> "a"^-) "f"(x)` is equal to f(a).


f(x) = |x| + |x − 1| at x = 1


f(x) = `{{:((2^(x + 2) - 16)/(4^x - 16)",",  "if"  x ≠ 2),("k"",",  "if"  x = 2):}` at x = 2


Find the values of a and b such that the function f defined by
f(x) = `{{:((x - 4)/(|x - 4|) + "a"",",  "if"  x < 4),("a" + "b"",",  "if"  x = 4),((x - 4)/(|x - 4|) + "b"",", "if"  x > 4):}`
is a continuous function at x = 4.


Given the function f(x) = `1/(x + 2)`. Find the points of discontinuity of the composite function y = f(f(x))


If f(x) = `x^2 sin  1/x` where x ≠ 0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is ______.


The value of k (k < 0) for which the function f defined as

f(x) = `{((1-cos"kx")/("x"sin"x")","  "x" ≠ 0),(1/2","  "x" = 0):}`

is continuous at x = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×