मराठी

If F is Defined by F (X) = X2, Find F'(2). - Mathematics

Advertisements
Advertisements

प्रश्न

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

थोडक्यात उत्तर

उत्तर

Given:  

\[f(x) = x^2\]

We know a  polynomial function is everywhere differentiable. Therefore 

\[f(x)\]  is differentiable at 
\[x = 2\]

\[f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{(2 + h )^2 - 2^2}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{(4 + h^2 + 4h) - 4}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{h (h + 4)}{h}\]
\[ \Rightarrow f'(2) = 4\]

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 1 | 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.


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):}`


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.

Show that 

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


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

 


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 value of the constant k so that the function 

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


If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\]  is continuous at x = 4, find ab.

 


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


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}k( x^2 + 2), \text{if} & x \leq 0 \\ 3x + 1 , \text{if} & x > 0\end{cases}\]


Discuss the continuity of the f(x) at the indicated points: 

(i) f(x) = | x | + | x − 1 | at x = 0, 1.


Prove that  \[f\left( x \right) = \begin{cases}\frac{x - \left| x \right|}{x}, & x \neq 0 \\ 2 , & x = 0\end{cases}\] is discontinuous at x = 0

 


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if }   x \neq 0 \\ 4 , & \text{ if }  x = 0\end{cases}\]

 


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].


Write the value of b for which \[f\left( x \right) = \begin{cases}5x - 4 & 0 < x \leq 1 \\ 4 x^2 + 3bx & 1 < x < 2\end{cases}\]  is continuous at x = 1.

 


The function 

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

 


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

 


If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\]  , then

 


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


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 values of the constants ab and c for which the function  \[f\left( x \right) = \begin{cases}\left( 1 + ax \right)^{1/x} , & x < 0 \\ b , & x = 0 \\ \frac{\left( x + c \right)^{1/3} - 1}{\left( x + 1 \right)^{1/2} - 1}, & x > 0\end{cases}\] may be continuous at x = 0, are

 


Discuss the continuity and differentiability of f (x) = e|x| .


The set of points where the function f (x) = x |x| is differentiable is 

 


Let \[f\left( x \right) = \begin{cases}\frac{1}{\left| x \right|} & for \left| x \right| \geq 1 \\ a x^2 + b & for \left| x \right| < 1\end{cases}\] If f (x) is continuous and differentiable at any point, then

 

 

 


Let f (x) = |sin x|. Then,


Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .


Find the value of k for which the function f (x ) =  \[\binom{\frac{x^2 + 3x - 10}{x - 2}, x \neq 2}{ k , x^2 }\] is continuous at x = 2 .

 
 

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


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


If f(x) = `{{:((x^3 + x^2 - 16x + 20)/(x - 2)^2",", x ≠ 2),("k"",", x = 2):}` is continuous at x = 2, find the value of k.


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 ______.


y = |x – 1| is a continuous function.


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


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


f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1


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


If the following function is continuous at x = 2 then the value of k will be ______.

f(x) = `{{:(2x + 1",", if x < 2),(                 k",", if x = 2),(3x - 1",", if x > 2):}`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×