Advertisements
Advertisements
Question
If \[f\left( x \right) = \left| \log_e |x| \right|\]
Options
f (x) is continuous and differentiable for all x in its domain
f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1
(x) is neither continuous nor differentiable at x = ± 1
none of these
Solution
(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1
We have,
\[f\left( x \right) = \left| \log_e |x| \right|\]
\[\text{We know that log function is defined for positive value} . \]
\[\text{Here,} \left| x \right| \text { is positive for all non zero x} . \]
\[\text{Therefore, domain of function is R} - \left\{ 0 \right\}\]
And we know that logarithmic function is continuous in its domain.
\[\left( \text { LHD at x } = - 1 \right) = \lim_{x \to - 1^-} \frac{f\left( x \right) - f\left( - 1 \right)}{x - \left( - 1 \right)}\]
\[ = \lim_{x \to - 1^-} \frac{\log_e \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left[ - \left( - 1 - h \right) \right]}{- 1 - h + 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 + h \right)}{- h}\]
\[ = - 1\]
\[\left( \text { RHD at x } = - 1 \right) = \lim_{x \to - 1^+} \frac{f\left( x \right) - f\left( - 1 \right)}{x - \left( - 1 \right)}\]
\[ = \lim_{x \to - 1^+} \frac{- \log_e \left( - x \right) - 0}{x + 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left[ - \left( - 1 + h \right) \right]}{- 1 + h + 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left( 1 - h \right)}{h}\]
\[ = {- \lim}_{h \to 0} \frac{\log_e \left( 1 - h \right)}{h}\]
\[ = - 1 \times - 1 = 1\]
\[\text { Here, LHD }\neq \text { RHD }\]
\[\text{Therefore, the given function is not differentiable at x} = - 1 .\]
\[ = \lim_{x \to 1^-} \frac{- \log_e \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{- \log_e \left[ \left( 1 - h \right) \right]}{1 - h - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 - h \right)}{h}\]
\[ = - 1\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - \left( 1 \right)}\]
\[ = \lim_{x \to 1^+} \frac{\log_e \left( x \right) - 0}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left[ \left( 1 + h \right) \right]}{1 + h - 1}\]
\[ = \lim_{h \to 0} \frac{\log_e \left( 1 + h \right)}{h}\]
\[ = 1\]
\[\text { Here, LHD } \neq \text { RHD }\]
\[\text{Therefore, the given function is not differentiable at x} = 1 .\]
APPEARS IN
RELATED QUESTIONS
Discuss the continuity of the function f, where f is defined by `f(x) = {(3, ","if 0 <= x <= 1),(4, ","if 1 < x < 3),(5, ","if 3 <= x <= 10):}`
Discuss the continuity of the function f, where f is defined by `f(x) = {(2x , ","if x < 0),(0, "," if 0 <= x <= 1),(4x, "," if x > 1):}`
Show that
Discuss the continuity of the following functions at the indicated point(s):
(ii) \[f\left( x \right) = \left\{ \begin{array}{l}x^2 \sin\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]
Discuss the continuity of the following functions at the indicated point(s):
Show that
\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{\tan 2x} , if x < 0 \\ \frac{3}{2} , if x = 0 \\ \frac{\log(1 + 3x)}{e^{2x} - 1} , if x > 0\end{cases}\text{is continuous at} x = 0\]
Discuss the continuity of \[f\left( x \right) = \begin{cases}2x - 1 & , x < 0 \\ 2x + 1 & , x \geq 0\end{cases} at x = 0\]
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
\[f\left( x \right) = \begin{cases}k x^2 , if & x \leq 2 \\ 3 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]
If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if } & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\] is continuous at x = 4, find a, b.
If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\]
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?
If \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if } x \neq 4 \\ k , & \text{ if } x = 4\end{cases}\] is continuous at x = 4, find k.
If f (x) = | x − a | ϕ (x), where ϕ (x) is continuous function, then
If \[f\left( x \right) = \left| \log_{10} x \right|\] then at x = 1
If \[f\left( x \right) = \begin{cases}\frac{\sin (a + 1) x + \sin x}{x} , & x < 0 \\ c , & x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{bx\sqrt{x}} , & x > 0\end{cases}\]is continuous at x = 0, then
If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\] , then
The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]
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
Define differentiability of a function at a point.
Is every continuous function differentiable?
If f (x) is differentiable at x = c, then write the value of
The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
If f is continuous at x = 0 then find f(0) where f(x) = `[5^x + 5^-x - 2]/x^2`, x ≠ 0
If the function f is continuous at x = 0
Where f(x) = 2`sqrt(x^3 + 1)` + a, for x < 0,
= `x^3 + a + b, for x > 0
and f (1) = 2, then find a and b.
If f(x) = `(e^(2x) - 1)/(ax)` . for x < 0 , a ≠ 0
= 1. for x = 0
= `(log(1 + 7x))/(bx)`. for x > 0 , b ≠ 0
is continuous at x = 0 . then find a and b
Examine the continuity off at x = 1, if
f (x) = 5x - 3 , for 0 ≤ x ≤ 1
= x2 + 1 , for 1 ≤ x ≤ 2
Examine the continuity of the followin function :
`{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "x=0`
Find the value of the constant k so that the function f defined below is continuous at x = 0, where f(x) = `{{:((1 - cos4x)/(8x^2)",", x ≠ 0),("k"",", 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 number of points at which the function f(x) = `1/(log|x|)` is discontinuous is ______.
A continuous function can have some points where limit does not exist.
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.
Given the function f(x) = `1/(x + 2)`. Find the points of discontinuity of the composite function y = f(f(x))
Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
