English

The Function F (X) = Sin−1 (Cos X) is (A) Discontinuous at X = 0 (B) Continuous at X = 0 (C) Differentiable at X = 0 (D) None of These - Mathematics

Advertisements
Advertisements

Question

The function f (x) = sin−1 (cos x) is

Options

  • discontinuous at x = 0

  • continuous at x = 0

  • differentiable at x = 0

  • none of these

MCQ
Advertisements

Solution

(b) continuous at x = 0 

Given:  

\[f(x) = \sin^{- 1} \left( \cos x \right) .\]

Continuity at x = 0: 

We have,
(LHL at x = 0) 

\[\lim_{x \to 0^-} f(x) \]
\[ = \lim_{h \to 0} \sin^{- 1} \left\{ \cos\left( 0 - h \right) \right\}\]
\[ = \lim_{h \to 0} \sin^{- 1} \left( \cos h \right)\]
\[ = \sin^{- 1} \left( 1 \right)\]
\[ = \frac{\pi}{2}\]

(RHL at x = 0)

\[\lim_{x \to 0^+} f\left( x \right)\]
\[ = \lim_{h \to 0} \sin^{- 1} \cos\left( 0 + h \right)\]
\[ = \lim_{h \to 0} \sin^{- 1} \left( \cos h \right)\]
\[ = \sin^{- 1} \left( 1 \right) \]
\[ = \frac{\pi}{2}\]

\[f(0) = \sin^{- 1} \left( \cos 0 \right) \]
\[ = \sin^{- 1} \left( 1 \right)\]
\[ = \frac{\pi}{2}\]

\[\lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( 0 - h \right) - \frac{\pi}{2}}{- h} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( - h \right) - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( \frac{\pi}{2} - h \right) \right\} - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{- h}{- h}\]
\[ = 1\]

RHD at x = 0

\[\lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( 0 + h \right) - \frac{\pi}{2}}{h} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - \frac{\pi}{2}}{h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( \frac{\pi}{2} - h \right) \right\} - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{- h}{h}\]
\[ = - 1\]

\[\therefore LHD \neq RHD\]

Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

shaalaa.com
  Is there an error in this question or solution?
Chapter 10: Differentiability - Exercise 10.4 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 10 Differentiability
Exercise 10.4 | Q 2 | Page 17

RELATED QUESTIONS

If f(x)= `{((sin(a+1)x+2sinx)/x,x<0),(2,x=0),((sqrt(1+bx)-1)/x,x>0):}`

is continuous at x = 0, then find the values of a and b.


A function f(x) is defined as,

\[f\left( x \right) = \begin{cases}\frac{x^2 - x - 6}{x - 3}; if & x \neq 3 \\ 5 ; if & x = 3\end{cases}\]  Show that f(x) is continuous that x = 3.

If \[f\left( x \right) = \begin{cases}e^{1/x} , if & x \neq 0 \\ 1 , if & x = 0\end{cases}\] find whether f is continuous at x = 0.


Show that

\[f\left( x \right)\] = \begin{cases}\frac{x - \left| x \right|}{2}, when & x \neq 0 \\ 2 , when & x = 0\end{cases}

is discontinuous at x = 0.

 

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \binom{\left| x - a \right|\sin\left( \frac{1}{x - a} \right), for x \neq a}{0, for x = a}at x = a\] 

Show that 

\[f\left( x \right) = \begin{cases}1 + x^2 , if & 0 \leq x \leq 1 \\ 2 - x , if & x > 1\end{cases}\]


For what value of k is the following function continuous at x = 1? \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}, & x \neq 1 \\ k , & x = 1\end{cases}\]


Determine the value of the constant k so that the function 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{x^2 - 3x + 2}{x - 1}, if & x \neq 1 \\ k , if & x = 1\end{array}\text{is continuous at x} = 1 \right.\] 


If   \[f\left( x \right) = \begin{cases}\frac{2^{x + 2} - 16}{4^x - 16}, \text{ if } & x \neq 2 \\ k , \text{ if }  & x = 2\end{cases}\]  is continuous at x = 2, find k.


Discuss the continuity of the f(x) at the indicated points:  f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

 

Prove that
\[f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}\] is everywhere continuous.

 


Determine if \[f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}\] is a continuous function?

 


Given the function  
\[f\left( x \right) = \frac{1}{x + 2}\] . Find the points of discontinuity of the function f(f(x)).

The function 

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

 


Let f (x) = | x | + | x − 1|, then


The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


If  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (ab) for which f (x) cannot be continuous at x = 1, is

 


The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]  


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


Show that f(x) = |x − 2| is continuous but not differentiable at x = 2. 


Discuss the continuity and differentiability of f (x) = |log |x||.


Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.


Discuss the continuity of f at x = 1 ,
Where f(x) = `(3 - sqrt(2x + 7))/(x - 1)` for x = ≠ 1
= `(-1)/3`   for x = 1


Discuss the continuity of f at x = 1
Where f(X) = `[ 3 - sqrt ( 2x + 7 ) / ( x - 1 )]`           For x ≠ 1
                    = `-1/3`                                                 For x = 1


Find the value of 'k' if the function 
f(x) = `(tan 7x)/(2x)`,                   for x ≠ 0.
      = k                                        for x = 0.
is continuous at x = 0.


If f (x) = `(1 - "sin x")/(pi - "2x")^2` , for x ≠ `pi/2` is continuous at x = `pi/4` , then find `"f"(pi/2) .`


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous at x = 0.


If f(x) = `(sqrt(2) cos x - 1)/(cot x - 1), x ≠ pi/4` find the value of `"f"(pi/4)`  so that f (x) becomes continuous at x = `pi/4`


The function given by f (x) = tanx is discontinuous on the set ______.


The function f(x) = |x| + |x – 1| is ______.


Examine the continuity of the function f(x) = x3 + 2x2 – 1 at x = 1


f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0


f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1


Examine the differentiability of f, where f is defined by
f(x) = `{{:(1 + x",",  "if"  x ≤ 2),(5 - x",",  "if"  x > 2):}` at x = 2


Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.


If f is continuous on its domain D, then |f| is also continuous on D.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×