हिंदी

Show that the function defined by f(x) = |cos x| is a continuous function. - Mathematics

Advertisements
Advertisements

प्रश्न

Show that the function defined by f(x) = |cos x| is a continuous function.

योग
Advertisements

उत्तर

It is known that f(x) = |cos x|

Let x = c ∈ R

`lim_(x -> c)` f(x) = `lim_(x -> c)` |cos x| = |cos c|

f(c) = |cos c|

Hence, f is a continuous function on x = c ∈ R.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १६१]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.1 | Q 32 | पृष्ठ १६१

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

If f (x) is continuous on [–4, 2] defined as 

f (x) = 6b – 3ax, for -4 ≤ x < –2
       = 4x + 1,    for –2 ≤ x ≤ 2

Show that a + b =`-7/6`


Find the values of a and b such that the function defined by f(x) = `{(5", if"  x <= 2),(ax +b", if"  2 < x < 10),(21", if"  x >= 10):}` is a continuous function.


Examine the continuity of the function  

\[f\left( x \right) = \left\{ \begin{array}{l}3x - 2, & x \leq 0 \\ x + 1 , & x > 0\end{array}at x = 0 \right.\]

Also sketch the graph of this function.


Determine the value of the constant k so that the function

\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0 .\]

 


Extend the definition of the following by continuity 

\[f\left( x \right) = \frac{1 - \cos7 (x - \pi)}{5 (x - \pi )^2}\]  at the point x = π.

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if }  & x = 0\end{cases}\] at x = 0


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}(x - 1)\tan\frac{\pi  x}{2}, \text{ if } & x \neq 1 \\ k , if & x = 1\end{cases}\] at x = 1at x = 1


If  \[f\left( x \right) = \begin{cases}2 x^2 + k, &\text{ if }  x \geq 0 \\ - 2 x^2 + k, & \text{ if }  x < 0\end{cases}\]  then what should be the value of k so that f(x) is continuous at x = 0.

 


Prove that the function \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & x < 0 \\ x + 1, & x \geq 0\end{cases}\]  is everywhere continuous.

 


Discuss the continuity of the function  

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

Find the points of discontinuity, if any, of the following functions: 

\[f\left( x \right) = \begin{cases}x^3 - x^2 + 2x - 2, & \text{ if }x \neq 1 \\ 4 , & \text{ if } x = 1\end{cases}\]

 


Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if }  x = 2\end{cases}\]


In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if  }  x \leq 2 \\ x - 1, & \text{ if }  x > 2\end{cases}\]


Discuss the continuity of f(x) = sin | x |.


Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.


What happens to a function f (x) at x = a, if  

\[\lim_{x \to a}\] f (x) = f (a)?

Determine whether \[f\left( x \right) = \binom{\frac{\sin x^2}{x}, x \neq 0}{0, x = 0}\]  is continuous at x = 0 or not.

 


If  \[f\left( x \right) = \binom{\frac{1 - \cos x}{x^2}, x \neq 0}{k, x = 0}\]  is continuous at x = 0, find k


 then f (x) is continuous for all
\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - x \right|}{x^2 - x}, & x \neq 0, 1 \\ 1 , & x = 0 \\ - 1 , & x = 1\end{cases}\]  then f (x) is continuous for all

If  \[f\left( x \right) = \begin{cases}\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] and f (x) is continuous at x = 0, then the value of k is


The function 

\[f\left( x \right) = \begin{cases}x^2 /a , & 0 \leq x < 1 \\ a , & 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \sqrt{2} \leq x < \infty\end{cases}\]is continuous for 0 ≤ x < ∞, then the most suitable values of a and b are

 


If  \[f\left( x \right) = \frac{1 - \sin x}{\left( \pi - 2x \right)^2},\] when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x= π/2, where λ =


The value of a for which the function \[f\left( x \right) = \begin{cases}\frac{\left( 4^x - 1 \right)^3}{\sin\left( x/a \right) \log \left\{ \left( 1 + x^2 /3 \right) \right\}}, & x \neq 0 \\ 12 \left( \log 4 \right)^3 , & x = 0\end{cases}\]may be continuous at x = 0 is

 


If  \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]then the value of the function at = 0, so that the function is continuous at x = 0, is

 


The value of a for which the function \[f\left( x \right) = \begin{cases}5x - 4 , & \text{ if } 0 < x \leq 1 \\ 4 x^2 + 3ax, & \text{ if } 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 


Find the values of a and b so that the function

\[f\left( x \right)\begin{cases}x^2 + 3x + a, & \text { if } x \leq 1 \\ bx + 2 , &\text {  if } x > 1\end{cases}\] is differentiable at each x ∈ R.

If is defined by  \[f\left( x \right) = x^2 - 4x + 7\] , show that \[f'\left( 5 \right) = 2f'\left( \frac{7}{2} \right)\] 


The function f (x) = |cos x| is


The function f (x) = x − [x], where [⋅] denotes the greatest integer function is


Let f (x) = |cos x|. Then,


If f(x) = 2x and g(x) = `x^2/2 + 1`, then which of the following can be a discontinuous function ______.


Let f(x) = |sin x|. Then ______.


The point(s), at which the function f given by f(x) = `{("x"/|"x"|","  "x" < 0),(-1","  "x" ≥ 0):}` is continuous, is/are:


If `f`: R → {0, 1} is a continuous surjection map then `f^(-1) (0) ∩ f^(-1) (1)` is:


The function f(x) = x2 – sin x + 5 is continuous at x =


For what value of `k` the following function is continuous at the indicated point

`f(x) = {{:(kx + 1",", if x ≤ pi),(cos x",", if x > pi):}` at = `pi`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×