Advertisements
Advertisements
प्रश्न
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if } x \neq 0 \\ 3k , & \text{ if } x = 0\end{cases}\]
Advertisements
उत्तर
Given:
\[ \Rightarrow \lim_{x \to 0} \frac{2\sin 2x}{2 \times 5x} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{5} \lim_{x \to 0} \frac{\sin 2x}{2x} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{5} = 3k\]
\[ \Rightarrow k = \frac{2}{15}\]
APPEARS IN
संबंधित प्रश्न
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.
Discuss the continuity of the following function:
f(x) = sin x × cos x
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<= pi),(cos x", if" x > pi):}` at x = π
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.
Determine the value of the constant k so that the function
\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0 .\]
Find the values of a so that the function
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.
Find the points of discontinuity, if any, of the following functions:
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}\]
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.
If \[f\left( x \right) = \begin{cases}\frac{x}{\sin 3x}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then write the value of k.
If \[f\left( x \right) = \binom{\frac{1 - \cos x}{x^2}, x \neq 0}{k, x = 0}\] is continuous at x = 0, find k.
Determine the value of the constant 'k' so that function f
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 =
The function \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\]
The function
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 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 \[f\left( x \right) = \begin{cases}\frac{1 - \cos 10x}{x^2} , & x < 0 \\ a , & x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + \sqrt{x}} - 25}, & x > 0\end{cases}\] then the value of a so that f (x) may be continuous at x = 0, is
If \[f \left( x \right) = \sqrt{x^2 + 9}\] , write the value of
The function f (x) = |cos x| is
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
The function f(x) = `"e"^|x|` 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
A real value of x satisfies `((3 - 4ix)/(3 + 4ix))` = α – iβ (α, β ∈ R), if α2 + β2 is equal to
The function f(x) = 5x – 3 is continuous at x =
The function f(x) = x2 – sin x + 5 is continuous at x =
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
