मराठी

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

Advertisements
Advertisements

प्रश्न

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

 
बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Continuity - Exercise 9.2 [पृष्ठ ३७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.2 | Q 15 | पृष्ठ ३७

व्हिडिओ 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`


For what value of λ is the function defined by f(x) = `{(λ(x^2 - 2x)", if"  x <= 0),(4x+ 1", if"  x > 0):}` continuous at x = 0? What about continuity at x = 1?


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 +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.


Determine the value of the constant k so that the function

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

 


Find the values of a so that the function 

\[f\left( x \right) = \begin{cases}ax + 5, if & x \leq 2 \\ x - 1 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]

Find the value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]


Let  \[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}\] x ≠ 0. Find the value of f at x = 0 so that f becomes continuous 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 \[f\left( x \right) = \begin{cases}\frac{x^2}{2}, & \text{ if } 0 \leq x \leq 1 \\ 2 x^2 - 3x + \frac{3}{2}, & \text P{ \text{ if }  }  1 < x \leq 2\end{cases}\]. Show that f is continuous at 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.

 


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:  \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if }  x \neq 0 \\ 3k , & \text{ if  } x = 0\end{cases}\] 


In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]


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


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


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

\[\lim_{x \to a}\] f (x) = f (a)?

If \[f\left( x \right) = \begin{cases}\frac{1 - \sin x}{\left( \pi - 2x \right)^2} . \frac{\log \sin x}{\log\left( 1 + \pi^2 - 4\pi x + 4 x^2 \right)}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k =

 


If f (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to 


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

 


The function 

\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & x \neq 0 \\ \frac{k}{2} , & x = 0\end{cases}\]  is continuous at x = 0, then k =

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


If the function f (x) defined by  \[f\left( x \right) = \begin{cases}\frac{\log \left( 1 + 3x \right) - \log \left( 1 - 2x \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then k =

 


If \[f\left( x \right) = \begin{cases}\frac{1 - \cos 10x}{x^2} , & x < 0 \\ a , & x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + \sqrt{x}} - 25}, & x > 0\end{cases}\] then the value of a so that f (x) may be continuous at x = 0, is 


Find the values of a and b, if the function f defined by 

\[f\left( x \right) = \begin{cases}x^2 + 3x + a & , & x \leqslant 1 \\ bx + 2 & , & x > 1\end{cases}\] is differentiable at = 1.

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


If \[f\left( x \right) = \begin{cases}\frac{\left| x + 2 \right|}{\tan^{- 1} \left( x + 2 \right)} & , x \neq - 2 \\ 2 & , x = - 2\end{cases}\]  then f (x) 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 ______.


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


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


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


The function f(x) = 5x – 3 is continuous at x =


What is the values of' 'k' so that the function 'f' is continuous at the indicated point


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×