मराठी

Find Whether the Function is Differentiable at X = 1 and X = 2 F ( X ) = ⎧ ⎨ ⎩ X X ≤ 1 − X − 2 + 3 X − X 2 ≤ X ≤ 2 X > 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

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

उत्तर

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

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

\[\text { Now }, \]
\[\text { LHL } = \lim_{x \to 1^-} f'\left( x \right) = \lim_{x \to 1^-} 1 = 1\]
\[\text { RHL } = \lim_{x \to 1^+} f'\left( x \right) = \lim_{x \to 1^+} - 1 = - 1\]
\[\text { Since , at x } = 1, \text { LHL} \neq \text { RHL }\]
\[\text { Hence }, f\left( x \right) \text { is not differentiable at } x = 1\]
\[\text { Again }, \]
\[\text { LHL }= \lim_{x \to 2^-} f'\left( x \right) = \lim_{x \to 2^-} - 1 = - 1\]
\[\text { RHL }= \lim_{x \to 2^+} f'\left( x \right) = \lim_{x \to 2^+} 3 - 2x = 3 - 4 = - 1\]
\[\text { Since , at x = 2, LHL = RHL}\]
\[\text { Hence,} f\left( x \right) \text { is differentiable at } x = 2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Differentiability - Exercise 10.1 [पृष्ठ १०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 10 Differentiability
Exercise 10.1 | Q 6 | पृष्ठ १०

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Examine the following function for continuity:

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


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

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

 


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;  \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5


Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.

 

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


The function 

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

 


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


If the function \[f\left( x \right) = \begin{cases}\left( \cos x \right)^{1/x} , & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then the value of k is


Let f (x) = | x | + | x − 1|, then


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

 


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


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


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 

\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.


If f is defined by f (x) = x2, find f'(2).


Discuss the continuity and differentiability of 

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

Define differentiability of a function at a point.

 

If f (x) is differentiable at x = c, then write the value of 

\[\lim_{x \to c} f \left( x \right)\]

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


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


The total cost C for producing x units is Rs (x2 + 60x + 50) and the price is Rs (180 - x) per unit. For how many units the profit is maximum.


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 Y = tan-1 `[(cos 2x - sin 2x)/(sin2x + cos 2x)]` then find `(dy)/(dx)`


Discuss the continuity of the function f at x = 0, where
f(x) = `(5^x + 5^-x - 2)/(cos2x - cos6x),` for x ≠ 0
      = `1/8(log 5)^2,`  for 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?


The number of points at which the function f(x) = `1/(x - [x])` is not continuous is ______.


The function given by f (x) = tanx is discontinuous on the set ______.


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


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


f(x) = |x| + |x − 1| at x = 1


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


Prove that the function f defined by 
f(x) = `{{:(x/(|x| + 2x^2)",",  x ≠ 0),("k",  x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.


Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.


`lim_("x" -> "x" //4) ("cos x - sin x")/("x"- "x" /4)`  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×