English

The Value of F (0), So that the Function F ( X ) = ( 27 − 2 X ) 1 / 3 − 3 9 − 3 ( 243 + 5 X ) 1 / 5 ( X ≠ 0 ) is Continuous, is Given by (A) 2 3

Advertisements
Advertisements

Question

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 

Options

  • \[\frac{2}{3}\]

  • 6

  • 2

  • 4

MCQ
Advertisements

Solution

For f(x) to be continuous at x = 0, we must have 

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - 3}{9 - 3 \left( 243 + 5x \right)^\frac{1}{5}}\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{3\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}\]
\[ = \frac{1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}{x}}\]
\[ = \frac{- 1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{- 2x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{5x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{27 - 2x - 27}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{243 + 5x - 243}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times {27}^\frac{- 2}{3}}{\frac{1}{5} \times {243}^\frac{- 4}{5}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{{27}^\frac{2}{3}}}{\frac{1}{5} \times \frac{1}{{243}^\frac{4}{5}}}\]
\[ = 2\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Continuity - Exercise 9.4 [Page 44]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 8 Continuity
Exercise 9.4 | Q 18 | Page 44

RELATED QUESTIONS

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 <= 5),(3x - 5", if"  x > 5):}` at x = 5


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 (x2) 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 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: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if }  x = 2\end{cases}\]


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


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.


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

 


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  \[f\left( x \right) = \begin{cases}\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] and f (x) is continuous at x = 0, then the value of k is


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 value of a for which the function \[f\left( x \right) = \begin{cases}\frac{\left( 4^x - 1 \right)^3}{\sin\left( x/a \right) \log \left\{ \left( 1 + x^2 /3 \right) \right\}}, & x \neq 0 \\ 12 \left( \log 4 \right)^3 , & x = 0\end{cases}\]may be continuous at x = 0 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

 


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


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


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


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


`lim_("x"-> pi) (1 + "cos"^2 "x")/("x" - pi)^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) = ____________.


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


The value of f(0) for the function `f(x) = 1/x[log(1 + x) - log(1 - x)]` to be continuous at x = 0 should be


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


The function f(x) = x2 – sin x + 5 is continuous at x =


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 ______.


Discuss the continuity of the following function:

f(x) = sin x + cos x


If \[g\] is continuous and \[g(x)\neq 0\], which function is continuous?


If \[f\] is continuous, which scalar multiple is continuous?


Which conditions guarantee that \[f\circ g\] is continuous at \[c\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×