हिंदी

Let F ( X ) = Log ( 1 + X a ) − Log ( 1 − X B ) X X ≠ 0. Find the Value of F at X = 0 So that F Becomes Continuous at X = 0.

Advertisements
Advertisements

प्रश्न

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.

 

योग
Advertisements

उत्तर

Given: 

\[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}, x \neq 0\]

If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x} \right) = f\left( 0 \right)\]
\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{ax}{a}} - \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{bx}{b}} \right) = f\left( 0 \right)\]
\[\Rightarrow \frac{1}{a} \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{x}{a}} \right) - \left( - \frac{1}{b} \right) \lim_{x \to 0} \left( \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{- x}{b}} \right) = f\left( 0 \right)\]

\[\Rightarrow \frac{1}{a} \times 1 - \left( - \frac{1}{b} \right) \times 1 = f\left( 0 \right) \left[ \text{ Using } : \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 \right]\]
\[ \Rightarrow \frac{1}{a} + \frac{1}{b} = f\left( 0 \right)\]
\[ \Rightarrow \frac{a + b}{ab} = f\left( 0 \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Continuity - Exercise 9.1 [पृष्ठ १९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.1 | Q 30 | पृष्ठ १९

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.


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


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


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.

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.


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


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: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if  }  x \leq 2 \\ x - 1, & \text{ if }  x > 2\end{cases}\]


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.


Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x


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


If \[f\left( x \right) = \begin{cases}\frac{\sin^{- 1} x}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, write the value of k.


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


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


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

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

The function f (x) = |cos x| is


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


Let f (x) = |cos x|. Then,


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


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


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


The function f(x) = x |x| is ______.


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


Which statement about \[\sin x\] and \[\cos x\] is correct?


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×