हिंदी

If F ( X ) = X 2 + X 2 1 + X 2 + X 2 ( 1 + X 2 ) + . . . + X 2 ( 1 + X 2 ) + . . . . , Then at X = 0, F (X) (A) Has No Limit (B) is Discontinuous (C) is Continuous but Not Differentiable - Mathematics

Advertisements
Advertisements

प्रश्न

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)

विकल्प

  • has no limit

  • is discontinuous

  • is continuous but not differentiable

  • is differentiable

MCQ
Advertisements

उत्तर

(b) is discontinuous 

\[\text{We have}, \]
\[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . , \]
\[\text{When x = 0 then } x^2 = 0\]
\[ \text { and } \frac{x^2}{1 + x^2} = 0\]
\[ \therefore f\left( 0 \right) = 0 + 0 + 0 + 0 . . . . . . . \]
\[ \Rightarrow f\left( 0 \right) = 0\]
\[\text { When, x } \neq 0\]
\[\text{Then,} x^2 > 0\]
\[\text { and }1 + x^2 > x^2 \]
\[ \Rightarrow 0 < \frac{x^2}{1 + x^2} < 1\]
\[ \therefore \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . , \right)\]
\[ = \lim_{x \to 0} \left[ x^2 \left( 1 + \frac{1}{1 + x^2} + \frac{1}{\left( 1 + x^2 \right)} + . . . + \frac{1}{\left( 1 + x^2 \right)} + . . . . , \right) \right]\]
\[ = \lim_{x \to 0} \left[ x^2 \left( \frac{1}{1 - \frac{1}{1 + x^2}} \right) \right] \left[ \text{Sum of infinite series where}, r = \frac{1}{1 + x^2} \right]\]
\[ = \lim_{x \to 0} \left[ x^2 \left( \frac{1 + x^2}{x^2} \right) \right]\]
\[ = \lim_{x \to 0} \left( 1 + x^2 \right)\]
\[ = 1\]
\[ \therefore \lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)\]
\[ \therefore f\left( x \right) \text { is discontinuous at } x = 0\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Differentiability - Exercise 10.4 [पृष्ठ १८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 10 Differentiability
Exercise 10.4 | Q 10 | पृष्ठ १८

वीडियो 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.


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


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

 

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): 

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


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

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

Discuss the continuity of \[f\left( x \right) = \begin{cases}2x - 1 & , x < 0 \\ 2x + 1 & , x \geq 0\end{cases} at x = 0\]


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.

 


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


If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\] 

 


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

 


In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}5 , & \text{ if }  & x \leq 2 \\ ax + b, & \text{ if } & 2 < x < 10 \\ 21 , & \text{ if }  & x \geq 10\end{cases}\]


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

The function 

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

 


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 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


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


The value of k which makes \[f\left( x \right) = \begin{cases}\sin\frac{1}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]    continuous at x = 0, is

 


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


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 number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.


The function f (x) =  |cos x| is


If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is


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


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous at 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?


y = |x – 1| is a continuous function.


f(x) = `{{:(|x - 4|/(2(x - 4))",", "if"  x ≠ 4),(0",", "if"  x = 4):}` at x = 4


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


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.


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.


Examine the differentiability of f, where f is defined by
f(x) = `{{:(x[x]",",  "if"  0 ≤ x < 2),((x - 1)x",",  "if"  2 ≤ x < 3):}` at x = 2


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×