Advertisements
Advertisements
Question
The function f (x) = |cos x| is
Options
differentiable at x = (2n + 1) π/2, n ∈ Z
continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
neither differentiable nor continuous at x = n ∈ Z
none of these
Solution
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
We have,
`⇒f(x) = {(cosx , 2npile x<(4n +1)pi/2),(0, x = (4n + 1)pi/2),(-cos x , (4n+1)pi/2 < x<(4n + 3)pi/2),(0, x = (4n +3)pi/2),(cos x , (4n + 3)pi/2 < xle (2n + 2)pi):}`\[\text{When, x is in first quadrant}, i . e . 2n\pi \leq x < \left( 4n + 1 \right)\frac{\pi}{2} , \text { we have }\]
\[ f\left( x \right) = \text{cos x which being a trigonometrical function is continuous and differentiable in} \left( 2n\pi, \left( 4n + 1 \right)\frac{\pi}{2} \right)\]
\[\text{When, x is in second quadrant or in third quadrant}, i . e . , \left( 4n + 1 \right)\frac{\pi}{2} < x < \left( 4n + 3 \right)\frac{\pi}{2} , \text { we have }\]
\[ f\left( x \right) = - \text{cos x which being a trigonometrical function is continuous and differentiable in} \left( \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2} \right)\]
\[\text{When, x is in fourth quadrant}, i . e . , \left( 4n + 3 \right)\frac{\pi}{2} < x \leq \left( 2n + 2 \right)\pi , \text { we have }\]
\[ f\left( x \right) = \text { cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 3 \right)\frac{\pi}{2}, \left( 2n + 2 \right)\pi \right)\]
\[\text { Thus possible point of non - differentiability of } f\left( x \right)\text { are x } = \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text { Now, LHD } \left[ at x = \left( 4n + 1 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{\cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{- \sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = - 1\]
\[\text { And RHD } \left( at x = \left( 4n + 1 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{- \cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = 1\]
\[ \therefore \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So } f\left( x \right)\text { is not differentiable at x }= \left( 4n + 1 \right)\frac{\pi}{2}\]
\[\text { Now, LHD }\left[\text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{- \cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = 1\]
\[ \text { And RHD } \left(\text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{\cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{- \sin x}{1 - 0} \left[ \text{By L'Hospital rule} \right]\]
\[ = - 1\]
\[ \therefore \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So } f\left( x \right) \text{is not differentiable at x} = \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text{Therefore}, f\left( x \right)\text { is neither differentiable at} \left( 4n + 1 \right)\frac{\pi}{2}\text { nor at } \left( 4n + 3 \right)\frac{\pi}{2}\]
\[i . e . f\left( x \right) \text{is not differentiable at odd multiples of} \frac{\pi}{2}\]
\[i . e . f\left( x \right) \text{is not differentiable at x} = \left( 2n + 1 \right)\frac{\pi}{2}\]
\[\text{Therefore, f(x) is everywhere continuous but not differentiable at} \left( 2n + 1 \right)\frac{\pi}{2} .\]
APPEARS IN
RELATED QUESTIONS
If 'f' is continuous at x = 0, then find f(0).
`f(x)=(15^x-3^x-5^x+1)/(xtanx) , x!=0`
Examine the following function for continuity:
f(x) = | x – 5|
Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`
A function f(x) is defined as
Show that f(x) is continuous at x = 3
Find the value of 'a' for which the function f defined by
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.
Find the value of k for which \[f\left( x \right) = \begin{cases}\frac{1 - \cos 4x}{8 x^2}, \text{ when} & x \neq 0 \\ k ,\text{ when } & x = 0\end{cases}\] is continuous at x = 0;
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5
For what value of k is the following function continuous at x = 2?
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if } 0 \leq x \leq 1\end{cases}\]
Find f (0), so that \[f\left( x \right) = \frac{x}{1 - \sqrt{1 - x}}\] becomes continuous at x = 0.
Write the value of b for which \[f\left( x \right) = \begin{cases}5x - 4 & 0 < x \leq 1 \\ 4 x^2 + 3bx & 1 < x < 2\end{cases}\] is continuous at x = 1.
The function
If \[f\left( x \right) = \left| \log_{10} x \right|\] then at x = 1
The values of the constants a, b and c for which the function \[f\left( x \right) = \begin{cases}\left( 1 + ax \right)^{1/x} , & x < 0 \\ b , & x = 0 \\ \frac{\left( x + c \right)^{1/3} - 1}{\left( x + 1 \right)^{1/2} - 1}, & x > 0\end{cases}\] may be continuous at x = 0, are
The points of discontinuity of the function
\[f\left( x \right) = \begin{cases}2\sqrt{x} , & 0 \leq x \leq 1 \\ 4 - 2x , & 1 < x < \frac{5}{2} \\ 2x - 7 , & \frac{5}{2} \leq x \leq 4\end{cases}\text{ is } \left( \text{ are }\right)\]
If \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at \[x = \frac{\pi}{2}\], if
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) = x1/3 is not differentiable at x = 0.
Show that \[f\left( x \right) =\]`{(12x, -,13, if , x≤3),(2x^2, +,5, if x,>3):}` is differentiable at x = 3. Also, find f'(3).
Show that the function
\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.
Discuss the continuity and differentiability of f (x) = |log |x||.
Define differentiability of a function at a point.
Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.
Let f (x) = |x| and g (x) = |x3|, then
The set of points where the function f (x) = x |x| is differentiable is
The function f (x) = e−|x| is
Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is
Discuss the continuity of function f at x = 0.
Where f(X) = `[ [sqrt ( 4 + x ) - 2 ]/ ( 3x )]`, For x ≠ 0
= `1/12`, For x = 0
The probability distribution function of continuous random variable X is given by
f( x ) = `x/4`, 0 < x < 2
= 0, Otherwise
Find P( x ≤ 1)
Discuss the continuity of the function `f(x) = (3 - sqrt(2x + 7))/(x - 1)` for x ≠ 1
= `-1/3` for x = 1, at x = 1
Discuss the continuity of the function f at x = 0, where
f(x) = `(5^x + 5^-x - 2)/(cos2x - cos6x),` for x ≠ 0
= `1/8(log 5)^2,` for x = 0
The set of points where the functions f given by f(x) = |x – 3| cosx is differentiable is ______.
f(x) = `{{:((1 - cos 2x)/x^2",", "if" x ≠ 0),(5",", "if" x = 0):}` at x = 0
Prove that the function f defined by
f(x) = `{{:(x/(|x| + 2x^2)",", x ≠ 0),("k", x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.
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
If f(x) = `x^2 sin 1/x` where x ≠ 0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is ______.
If f is continuous on its domain D, then |f| is also continuous on D.
Given functions `"f"("x") = ("x"^2 - 4)/("x" - 2) "and g"("x") = "x" + 2, "x" le "R"`. Then which of the following is correct?
