हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

(i) f(x) = cos (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cos c) = (cos c) = (cos c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(ii) f(x) = cosec (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x ->^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cosec c) = (cosec c) = (cosec c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(iii) f(x) = sec (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (sec c) = (sec c) = (sec c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(iv) f(x) = cot (x)

Let c be any real number such that (n − 1)π < x < nπ, where n represents an integer point.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cot c) = (cot c) = (cot c)

This statement is true; that is, f(x) is continuous at every point on the real number line between (n − 1)π and nπ.

Now if we consider c such that c = nπ, where n represents an integer point, then:

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ ±∞ = ±∞ = ±∞

That is, f(x) is continuous at every point on the real number line except at the nπ-type points.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १६०]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.1 | Q 22 | पृष्ठ १६०

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

f(x) = `{(kx + 1", if"  x <= 5),(3x - 5", if"  x > 5):}` at x = 5


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.


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 values of a and b so that the function f given by \[f\left( x \right) = \begin{cases}1 , & \text{ if } x \leq 3 \\ ax + b , & \text{ if } 3 < x < 5 \\ 7 , & \text{ if }  x \geq 5\end{cases}\] is continuous at x = 3 and x = 5.


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if }  x < 0 \\ 2x + 3, & x \geq 0\end{cases}\]


In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if }  x \neq 0 \\ 3k , & \text{ if  } x = 0\end{cases}\] 


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


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


The function f(x) is defined as follows: 

\[f\left( x \right) = \begin{cases}x^2 + ax + b , & 0 \leq x < 2 \\ 3x + 2 , & 2 \leq x \leq 4 \\ 2ax + 5b , & 4 < x \leq 8\end{cases}\]

If f is continuous on [0, 8], find the values of a and b.


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

 

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

\[\lim_{x \to a}\] f (x) = f (a)?

If \[f\left( x \right) = \begin{cases}\frac{x}{\sin 3x}, & x \neq 0 \\ k , & x = 0\end{cases}\]  is continuous at x = 0, then write the value of k.


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

 


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


The 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 =

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


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 


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


The function f (x) = |cos x| is


The function \[f\left( x \right) = \frac{\sin \left( \pi\left[ x - \pi \right] \right)}{4 + \left[ x \right]^2}\] , 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 \[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\] 

then at x = 0, f (x) is


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

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


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


Let `"f" ("x") = ("In" (1 + "ax") - "In" (1 - "bx"))/"x", "x" ne 0` If f (x) is continuous at x = 0, then f(0) = ____________.


A real value of x satisfies `((3 - 4ix)/(3 + 4ix))` = α – iβ (α, β ∈ R), if α2 + β2 is equal to


Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0


Discuss the continuity of the following function:

f(x) = sin x + cos x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×