Advertisements
Advertisements
प्रश्न
Show that the function defined by f(x) = |cos x| is a continuous function.
Advertisements
उत्तर
It is known that f(x) = |cos x|
Let x = c ∈ R
`lim_(x -> c)` f(x) = `lim_(x -> c)` |cos x| = |cos c|
f(c) = |cos c|
Hence, f is a continuous function on x = c ∈ R.
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 value of k so that the function f is continuous at the indicated point.
f(x) = `{((kcosx)/(pi-2x)", if" x != pi/2),(3", if" x = pi/2):}` at x = `"pi/2`
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.
Find the value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]
Extend the definition of the following by continuity
If \[f\left( x \right) = \begin{cases}\frac{x^2}{2}, & \text{ if } 0 \leq x \leq 1 \\ 2 x^2 - 3x + \frac{3}{2}, & \text P{ \text{ if } } 1 < x \leq 2\end{cases}\]. Show that f is continuous at x = 1.
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.
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.
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: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}\]
\[f\left( x \right) = \begin{cases}\frac{\left| x^2 - x \right|}{x^2 - x}, & x \neq 0, 1 \\ 1 , & x = 0 \\ - 1 , & x = 1\end{cases}\] then f (x) is continuous for all
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 =
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
If the function \[f\left( x \right) = \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x}\] is continuous at each point of its domain, then the value of f (0) is
Let \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}, x \neq \frac{\pi}{4} .\] The value which should be assigned to f (x) at \[x = \frac{\pi}{4},\]so that it is continuous everywhere is
If \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]then the value of the function at x = 0, so that the function is continuous at x = 0, is
Find the values of a and b, if the function f defined by
Let f (x) = |cos x|. Then,
The function f (x) = 1 + |cos x| is
Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if
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" -> 0) (1 - "cos" 4 "x")/"x"^2` is equal to ____________.
`lim_("x" -> 0) (1 - "cos x")/"x sin x"` 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 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
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
