हिंदी

Examine the differentiability of f, where f is defined byf(x) = ,if,if{1+x, if x≤25-x, if x>2 at x = 2 - Mathematics

Advertisements
Advertisements

प्रश्न

Examine the differentiability of f, where f is defined by
f(x) = `{{:(1 + x",",  "if"  x ≤ 2),(5 - x",",  "if"  x > 2):}` at x = 2

योग
Advertisements

उत्तर

f(x) is differentiable at x = 2 if Lf'(2) = Rf'(2)

∴ Lf'(2) = `lim_("h" -> 0) ("f"(2 - "h") - "f"(2))/(-"h")`

= `lim_("h" -> 0) ((1 + 2 - "h") - (1 + 2))/(-"h")`

= `lim_("h" -> 0) (3 - "h" - 3)/(-"h')`

= `(-"h")/(-"h")`

= 1

Rf'(2) = `lim_("h" -> 0) ("f"(2 + "h") - "f"(2))/"h"`

= `lim_("h" -> 0) ([5 - (2 + "h")] - (1 + 2))/"h"`

= `lim_("h" -> 0) (3 - "h" - 3)/"h"`

= `(-"h")/"h"`

= –1

So, Lf'(2) ≠ Rf'(2)

Hence, f(x) is not differentiable at x = 2.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ १०९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 22 | पृष्ठ १०९

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

Determine the value of 'k' for which the following function is continuous at x = 3

`f(x) = {(((x + 3)^2 - 36)/(x - 3),  x != 3), (k,  x = 3):}`


Examine the following function for continuity:

f(x) = x – 5


Examine the following function for continuity:

f(x) = `(x^2 - 25)/(x + 5)`, x ≠ −5


Discuss the continuity of the function f, where f is defined by:

f(x) = `{(2x", if"  x < 0),(0", if"  0 <= x <= 1),(4x", if"  x > 1):}`


If \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\]

Find whether f(x) is continuous at x = 0.

 

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 following function at the indicated point:

`f(x) = {{:(|x| cos (1/x)",", x ≠ 0),(0",", x = 0):} at  x = 0`


Show that 

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


Find the value of 'a' for which the function f defined by

\[f\left( x \right) = \begin{cases}a\sin\frac{\pi}{2}(x + 1), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\]  is continuous at x = 0.
 

 


Determine the value of the constant k so that the function

\[f\left( x \right) = \begin{cases}k x^2 , if & x \leq 2 \\ 3 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]


If  \[f\left( x \right) = \begin{cases}\frac{1 - \cos kx}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\text{is continuous at x} = 0, \text{ find } k .\]


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


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

Find all point of discontinuity of the function 

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

Define continuity of a function at a point.

 

If \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if }  x \neq 4 \\ k , & \text{ if }  x = 4\end{cases}\]  is continuous at x = 4, find k.


If \[f\left( x \right) = \left| \log_{10} x \right|\] then at x = 1


If  \[f\left( x \right) = \begin{cases}\frac{{36}^x - 9^x - 4^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, then k equals

 


Let f (x) = | x | + | x − 1|, 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

The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


The function  \[f\left( x \right) = \frac{x^3 + x^2 - 16x + 20}{x - 2}\] is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as ______.


If  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (ab) for which f (x) cannot be continuous at x = 1, is

 


Show that f(x) = x1/3 is not differentiable at x = 0.


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).


Discuss the continuity and differentiability of 

\[f\left( x \right) = \begin{cases}\left( x - c \right) \cos \left( \frac{1}{x - c} \right), & x \neq c \\ 0 , & x = c\end{cases}\]

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


If \[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,\] 

then at x = 0, f (x)


If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\]  then f (x) is 


Find k, if the function f is continuous at x = 0, where

`f(x)=[(e^x - 1)(sinx)]/x^2`,      for x ≠ 0

     = k                             ,        for x = 0


If the function f is continuous at x = 0 then find f(0),
where f(x) =  `[ cos 3x - cos x ]/x^2`, `x!=0`


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 continuity of the function f(x) = x3 + 2x2 – 1 at x = 1


f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2


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 ______.


`lim_("x" -> "x" //4) ("cos x - sin x")/("x"- "x" /4)`  is equal to ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×