English

If F (X) = (X + 1)Cot X Be Continuous at X = 0, Then F (0) is Equal to (A) 0 (B) 1/E (C) E (D) None of These - Mathematics

Advertisements
Advertisements

Question

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

Options

  • 0

  • 1/e

  • e

  • none of these

MCQ
Advertisements

Solution 1

Suppose 

\[f\left( x \right)\]  is continuous at  \[x = 0 .\]
Given:
\[f\left( x \right) = \left( x + 1 \right)^{\text{ cot } x}\]

\[\log f\left( x \right) = \left( \cot x \right) \left( \log \left( x + 1 \right) \right) \left[ \text{ Taking log on both sides }  \right]\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \cot x \right) \left( \log \left( x + 1 \right) \right)\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{\tan x} \right)\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \lim_{x \to 0} \frac{\left( \frac{\log \left( x + 1 \right)}{x} \right)}{\left( \frac{\tan x}{x} \right)}\]

\[ \Rightarrow \lim_{x \to 0} \log f\left( x \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)}\]

\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = \frac{\lim_{x \to 0} \left( \frac{\log \left( x + 1 \right)}{x} \right)}{\lim_{x \to 0} \left( \frac{\tan x}{x} \right)} \left[ \because f\left( x \right)\text{ is continuous at } x = 0 \right]\]

\[ \Rightarrow \log \left( \lim_{x \to 0} f\left( x \right) \right) = 1\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = e\]

\[ \Rightarrow f\left( 0 \right) = e \left[ \because f\left( x \right) \text{ is continuous at }  x = 0 \right]\]

shaalaa.com

Solution 2

For continuity at x = 0, we must have 

f(0) = `lim_("x"->0) "f"("x")`

`=lim_(x->0) ("x" + 1)^"cot x" = lim_(x->0) [(1 + "x")^(1/"x")]^("x cot x")`

`= lim_("x"->0)[(1 + "x")^(1/"x")]^(lim_("x"->0)("x"/("tan x"))) = "e"^1 = e`

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Continuity - Exercise 9.4 [Page 43]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 9 Continuity
Exercise 9.4 | Q 7 | Page 43

RELATED QUESTIONS

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


Show that the function defined by f(x) = cos (x2) is a continuous function.


Examine that sin |x| is a continuous function.


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.

 


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; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if }  & x = 0\end{cases}\] at x = 0


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.

 

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:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]


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].


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.


Show that f (x) = cos x2 is a continuous function.


Determine the value of the constant 'k' so that function 

\[\left( x \right) = \begin{cases}\frac{kx}{\left| x \right|}, &\text{ if }  x < 0 \\ 3 , & \text{ if } x \geq 0\end{cases}\]  is continuous at x  = 0  . 

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 =

 


The value of f (0), so that the function

\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by 


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 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.

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  \[f \left( x \right) = \sqrt{x^2 + 9}\] , write the value of

\[\lim_{x \to 4} \frac{f\left( x \right) - f\left( 4 \right)}{x - 4} .\]

If \[f\left( x \right) = a\left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3\] 


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


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


If f.g is continuous at x = a, then f and g are separately continuous at x = a.


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


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


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


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`


Find the values of `a` and ` b` such that the function by:

`f(x) = {{:(5",", if  x ≤ 2),(ax + b",", if 2 < x < 10),(21",", if x ≥ 10):}`

is a continuous function.


The value of ‘k’ for which the function f(x) = `{{:((1 - cos4x)/(8x^2)",",  if x ≠ 0),(k",",  if x = 0):}` is continuous at x = 0 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×