हिंदी

Examine the differentiability of the function f defined byf(x) = ,if,if,if{2x+3, if -3≤x<-2x+1, if -2≤x<0x+2, if 0≤x≤1 - Mathematics

Advertisements
Advertisements

प्रश्न

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):}`

योग
Advertisements

उत्तर

The only doubtful points for differentiability of f(x) are x = – 2 and x = 0.

Differentiability at x = – 2.

Now Lf'(–2) = `lim_("h" -> 0) ("f"(-2 + "h")  "f"(-2))/"h"`

= `lim_("h" -> 0^-) (2(-2 + "h") + 3 - (-2 + 1))/"h"`

= `lim_("h" -> 0^-)  (2"h")/"h"`

= `lim_("h" -> 0^-) 2`

= 2

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

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

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

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

= 1

Thus R f′(–2) ≠ Lf′(–2).

Therefore f is not differentiable at x = – 2.

Similarly, for differentiability at x = 0, we have

Lf'(0) = `lim_("h" -> 0^-) ("f"(0 + "h") - "f"(0))/"h"`

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

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

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

Which does not exist.

Hence f is not differentiable at x = 0.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Solved Examples [पृष्ठ १०१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Solved Examples | Q 22 | पृष्ठ १०१

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

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

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


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

(ii) \[f\left( x \right) = \left\{ \begin{array}{l}x^2 \sin\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]


Show that 

\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{\tan 2x} , if x < 0 \\ \frac{3}{2} , if x = 0 \\ \frac{\log(1 + 3x)}{e^{2x} - 1} , if x > 0\end{cases}\text{is continuous at} x = 0\]


Determine the value of the constant k so that the function 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{x^2 - 3x + 2}{x - 1}, if & x \neq 1 \\ k , if & x = 1\end{array}\text{is continuous at x} = 1 \right.\] 


For what value of k is the function 

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


Determine the value of the constant k so that the function

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


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  

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

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

 


Let  \[f\left( x \right) = \begin{cases}\frac{x^4 - 5 x^2 + 4}{\left| \left( x - 1 \right) \left( x - 2 \right) \right|}, & x \neq 1, 2 \\ 6 , & x = 1 \\ 12 , & x = 2\end{cases}\]. Then, f (x) is continuous on the set

 


If  \[f\left( x \right) = \begin{cases}\frac{\sin (a + 1) x + \sin x}{x} , & x < 0 \\ c , & x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{bx\sqrt{x}} , & x > 0\end{cases}\]is continuous at x = 0, then 


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

 


If  \[f\left( x \right) = \frac{1}{1 - x}\] , then the set of points discontinuity of the function f (f(f(x))) is


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


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

 


The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]  


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

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

If \[f\left( x \right) = \begin{cases}a x^2 - b, & \text { if }\left| x \right| < 1 \\ \frac{1}{\left| x \right|} , & \text { if }\left| x \right| \geq 1\end{cases}\]  is differentiable at x = 1, find a, b.


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


Let \[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]


Let f (x) = |x| and g (x) = |x3|, then


If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]


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

=`4x+3  "for"  x>3,  "at"  x=3` 


`f(x)=(x^2-9)/(x - 3)` is not defined at x = 3. what value should be assigned to f(3) for continuity of f(x) at = 3?


Discuss the continuity of f at x = 1
Where f(X) = `[ 3 - sqrt ( 2x + 7 ) / ( x - 1 )]`           For x ≠ 1
                    = `-1/3`                                                 For x = 1


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 f(x) = `(e^(2x) - 1)/(ax)` .                for x < 0 , a ≠ 0
         = 1.                             for x = 0
         = `(log(1 + 7x))/(bx)`.        for x > 0 , b ≠ 0
is continuous at x = 0 . then find a and b


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


Examine the continuity of the followin function : 

  `{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "x=0`   


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


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):}`


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


f(x) = `{{:(|x - "a"| sin  1/(x - "a")",",  "if"  x ≠ 0),(0",",  "if"  x = "a"):}` at x = a


f(x) = `{{:((sqrt(1 + "k"x) - sqrt(1 - "k"x))/x",",  "if" -1 ≤ x < 0),((2x + 1)/(x - 1)",",  "if"  0 ≤ x ≤ 1):}` at x = 0


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


If f(x) = `{{:("m"x + 1",",  "if"  x ≤ pi/2),(sin x + "n"",",  "If"  x > pi/2):}`, is continuous at x = `pi/2`, then ______.


`lim_("x" -> 0) (2  "sin x - sin"  2 "x")/"x"^3` is equal to ____________.


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×