Advertisements
Advertisements
प्रश्न
Show that the function defined by f(x) = |cos x| is a continuous function.
उत्तर
It is known that f(x) = `abs (cos x)`
Let x = c `in` on R.
`lim_(x -> c) f(x) = lim_(x -> c) abs (cos x) = abs (cos c)`
f(c) = `abs (cos c")`
Hence, f is a continuous function on x = c `in` R.
APPEARS IN
संबंधित प्रश्न
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`
Is the function defined by `f(x) = x^2 - sin x + 5` continuous at x = π?
Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Find the values 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 values of k so that the function f is continuous at the indicated point.
`f(x) = {(kx^2, "," if x<= 2),(3, "," if x > 2):} " at x" = 2`
Examine the continuity of the function
\[f\left( x \right) = \left\{ \begin{array}{l}3x - 2, & x \leq 0 \\ x + 1 , & x > 0\end{array}at x = 0 \right.\]
Also sketch the graph of this function.
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
Find the values of a and b so that the function f given by \[f\left( x \right) = \begin{cases}1 , & \text{ if } x \leq 3 \\ ax + b , & \text{ if } 3 < x < 5 \\ 7 , & \text{ if } x \geq 5\end{cases}\] is continuous at x = 3 and x = 5.
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: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if } x = 2\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}\]
The function \[f\left( x \right) = \begin{cases}x^2 /a , & \text{ if } 0 \leq x < 1 \\ a , & \text{ if } 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \text{ if } \sqrt{2} \leq x < \infty\end{cases}\] is continuous on (0, ∞), then find the most suitable values of a and b.
The function f(x) is defined as follows:
If f is continuous on [0, 8], find the values of a and b.
Show that f (x) = cos x2 is a continuous function.
Show that f (x) = | cos x | is a continuous function.
What happens to a function f (x) at x = a, if
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.
The value of a for which the function \[f\left( x \right) = \begin{cases}\frac{\left( 4^x - 1 \right)^3}{\sin\left( x/a \right) \log \left\{ \left( 1 + x^2 /3 \right) \right\}}, & x \neq 0 \\ 12 \left( \log 4 \right)^3 , & x = 0\end{cases}\]may be continuous at x = 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 the function f (x) defined by \[f\left( x \right) = \begin{cases}\frac{\log \left( 1 + 3x \right) - \log \left( 1 - 2x \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then k =
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
Let f (x) = |cos x|. Then,
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 ______.
Let f(x) = |sin x|. Then ______.
If f.g is continuous at x = a, then f and g are separately continuous at x = a.
`lim_("x" -> 0) (1 - "cos" 4 "x")/"x"^2` is equal to ____________.
If `f`: R → {0, 1} is a continuous surjection map then `f^(-1) (0) ∩ f^(-1) (1)` is:
A real value of x satisfies `((3 - 4ix)/(3 + 4ix))` = α – iβ (α, β ∈ R), if α2 + β2 is equal to
If `f(x) = {{:(-x^2",", "when" x ≤ 0),(5x - 4",", "when" 0 < x ≤ 1),(4x^2 - 3x",", "when" 1 < x < 2),(3x + 4",", "when" x ≥ 2):}`, then
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
Discuss the continuity of the following function:
f(x) = sin x – cos x