Advertisements
Advertisements
प्रश्न
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
पर्याय
continuous and differentiable
differentiable but not continuous
continuous but not differentiable
neither continuous nor differentiable
उत्तर
(a) continuous and differentiable
we have,
\[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\]
\[\text { Continuity at x } = 0\]
\[(\text { LHL at x }= 0) = {lim}_{x \to 0 -} f(x)\]
\[ = {lim}_{h \to 0} f(0 - h)\]
\[ = {lim}_{h \to 0} f( - h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos ( - h)}{( - h) \sin ( - h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos \ h \ {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos(0) . \frac{1}{0 \sin 0} \]
\[ = 0\]
\[(\text { RHL at x }= 0) = {lim}_{x \to 0^+} f(x)\]
\[ = {lim}_{h \to 0} f(0 + h)\]
\[ = {lim}_{h \to 0} f( h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos (h)}{(h) \sin (h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos\ h {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos 0 . \frac{1}{0 \sin 0}\]
\[ = 0\]
Hence, f(x)is continuous at x = 0.
For differentiability at x = 0
\[ = {lim}_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( - h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos( - h)}{- h \sin( - h)} - \frac{1}{2}}{- h}\]
\[ = \frac{1}{h} {lim}_{h \to 0} $\frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]
\[\text { RHD at x } = 0 ) = {lim}_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 + h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos (h)}{- h \sin(h)} - \frac{1}{2}}{- h}\]
\[ = - \frac{1}{h} {lim}_{h \to 0} \frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
A function f (x) is defined as
f (x) = x + a, x < 0
= x, 0 ≤x ≤ 1
= b- x, x ≥1
is continuous in its domain.
Find a + b.
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 relationship between a and b so that the function f defined by `f(x)= {(ax + 1, if x<= 3),(bx + 3, if x > 3):}` is continuous at x = 3.
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.
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.
Extend the definition of the following by continuity
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if } & x = 0\end{cases}\] at x = 0
Prove that the function \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & x < 0 \\ x + 1, & x \geq 0\end{cases}\] is everywhere continuous.
Discuss the continuity of the function
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if } x = 2\end{cases}\]
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: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}\]
If \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}\]
for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].
Show that f (x) = cos x2 is a continuous function.
What happens to a function f (x) at x = a, if
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.
Let \[f\left( x \right) = \left\{ \begin{array}\\ \frac{x - 4}{\left| x - 4 \right|} + a, & x < 4 \\ a + b , & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, & x > 4\end{array} . \right.\]Then, f (x) is continuous at x = 4 when
The function \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\]
\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by
The function
Find the values of a and b so that the function
Find the values of a and b, if the function f defined by
If f is defined by \[f\left( x \right) = x^2 - 4x + 7\] , show that \[f'\left( 5 \right) = 2f'\left( \frac{7}{2} \right)\]
Let f (x) = |cos x|. Then,
The function f (x) = 1 + |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
The function f(x) = `(4 - x^2)/(4x - x^3)` is ______.
`lim_("x" -> 0) (1 - "cos x")/"x sin x"` is equal to ____________.
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
Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0
The function f(x) = x2 – sin x + 5 is continuous at x =
What is the values of' 'k' so that the function 'f' is continuous at the indicated point
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
