हिंदी

Show that F (X) = | Cos X | is a Continuous Function.

Advertisements
Advertisements

प्रश्न

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

 
योग
Advertisements

उत्तर

The given function is `f(x)=|cos x|`

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where `g(x)=|x|  and  h(x)=cos  x`

`[∵(goh)(x)=g(h(x))=g(cos  x)=|cos  x|=f(x)]`

It has to be first proved that `g(x)=|x|  and  h(x)=cos x` are continuous functions.

`g(x)=|x| "  can be written as " `

`g(x)=[[-x,if x≤ 0],[x,if x ≥ 0]]`

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

`if c < 0 " then " g (c)= -c and lim\_(x->c)(-x)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x < 0.

Case II:

`" if c < 0   then "  g (c)= -c and lim\_(x->c)(-x)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x > 0.

Case III: 

`if  c = 0 , "  then " g(c)=g(0)=0`

`lim_(x->0^-)g(x)=lim_(x->0^-)(-x)=0`

`lim_(x->0^+)g(x)=lim_(x->0^+)(x)=g(0)`

`∴lim_(x->0^+)g(x)=lim_(x->0^+)(x)=g(0)`

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let be a real number.
Put x = c + h

If x → c, then h → 0

(c) = cos c

`lim_(x->0)h(x)=lim_(x->0) cos x`

                   `=lim_(k->0) cos (c+h)`

                   `=lim_(k->0)[cos  c  cos  h-sin  c sin h]`

                   `=lim_(k->0)cos  c cos 0 - sin  c sin 0`

                  `= cos  c xx1 - sin  cxx0`

                  `= cos  c`

`lim_(x->c)h(x)=h(c)`

So, h (x) = cos x is a continuous function.

It is known that for real valued functions and h,such that (h) is defined at x = c, if is continuous at x = and if is continuous at (c), then (g) is continuous at x = c.

Therefore, `f(x)=(goh)(x)=g(h(x))=g(cos x)=|cos x|` is a continuous function.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Continuity - Exercise 9.2 [पृष्ठ ३७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.2 | Q 15 | पृष्ठ ३७

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Discuss the continuity of the following function:

f(x) = sin x × cos x


Discuss the continuity of the cosine, cosecant, secant and cotangent functions.


Find the value of k so that the function f is continuous at the indicated point.

f(x) = `{(kx^2", if"  x<= 2),(3", if"  x > 2):}` at x = 2


Find the value of k so that the function f is continuous at the indicated point.

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


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.


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


If  \[f\left( x \right) = \frac{2x + 3\ \text{ sin }x}{3x + 2\ \text{ sin }  x}, x \neq 0\] If f(x) is continuous at x = 0, then find f (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


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


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if }  x = 0\end{cases}\]


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


In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}2 , & \text{ if }  x \leq 3 \\ ax + b, & \text{ if }  3 < x < 5 \\ 9 , & \text{ if }  x \geq 5\end{cases}\]


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


If \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}\]

for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].


If the function   \[f\left( x \right) = \frac{\sin 10x}{x}, x \neq 0\] is continuous at x = 0, find f (0).

 


 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  \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\] 


If the function  \[f\left( x \right) = \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x}\] is continuous at each point of its domain, then the value of f (0) is 


Let  \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}, x \neq \frac{\pi}{4} .\]  The value which should be assigned to f (x) at  \[x = \frac{\pi}{4},\]so that it is continuous everywhere 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 


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


Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if


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


The function f(x) = `(4 - x^2)/(4x - x^3)` is ______.


The function f(x) = `"e"^|x|` is ______.


`lim_("x" -> 0) (1 - "cos" 4 "x")/"x"^2` is equal to ____________.


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


If `f(x) = {{:(-x^2",", "when"  x ≤ 0),(5x - 4",", "when"  0 < x ≤ 1),(4x^2 - 3x",", "when"  1 < x < 2),(3x + 4",", "when"  x ≥ 2):}`, then


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

`f(x) = {{:(kx^2",", if x ≤ 2),(3",", if x > 2):}` at x = 2


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`


For \[\frac{f}{g}\] to be continuous at \[x=c\], which condition is required?


For \[f(x)=\frac{p(x)}{q(x)}\], where \[p(x)\] and \[q(x)\] are polynomial functions, where is \[f\] continuous?


Let \[g(x)=\sin x\] and \[h(x)=x^2\]. Which expression equals \[(g\circ h)(x)\]?


For \[f(x)=|1-x+|x||\], which functions give the representation \[f(x)=h(g(x))\]?


Why is \[g(x)=1-x+|x|\] continuous?


Why is \[f(x)=|1-x+|x||\] continuous for every real \[x\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×