Advertisements
Advertisements
प्रश्न
Determine the value of the constant 'k' so that function f
Advertisements
उत्तर
\[\text{ Given } , f(x) = \begin{cases}\frac{kx}{\left| x \right|} & , \text{ if } x < 0 \\ 3 & , \text{ if } x \geqslant 0\end{cases}\]
Since the function is continuous at x = 0, therefore,
\[\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)\]
\[ \Rightarrow \lim_{x \to 0} \frac{- kx}{x} = \lim_{x \to 0} 3 = 3\]
\[ \Rightarrow - k = 3\]
\[ \Rightarrow k = - 3\]
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`
Find the value 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
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
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 value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]
If \[f\left( x \right) = \frac{2x + 3\ \text{ sin }x}{3x + 2\ \text{ sin } x}, x \neq 0\] If f(x) is continuous at x = 0, then find f (0).
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}(x - 1)\tan\frac{\pi x}{2}, \text{ if } & x \neq 1 \\ k , if & x = 1\end{cases}\] at x = 1at x = 1
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.
Discuss the continuity of the function
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & \text{ if } x < 0 \\ 2x + 3, & x \geq 0\end{cases}\]
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{\sin x}{x} + \cos x, & \text{ if } x \neq 0 \\ 5 , & \text { if } x = 0\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}kx + 5, & \text{ if } x \leq 2 \\ x - 1, & \text{ if } x > 2\end{cases}\]
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.
Show that f (x) = | cos x | is a continuous function.
If the function \[f\left( x \right) = \frac{\sin 10x}{x}, x \neq 0\] is continuous at x = 0, find f (0).
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.
If \[f\left( x \right) = \frac{1 - \sin x}{\left( \pi - 2x \right)^2},\] when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x= π/2, where λ =
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
If \[f\left( x \right) = a\left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3\]
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
The function f(x) = `(4 - x^2)/(4x - x^3)` is ______.
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
If `f`: R → {0, 1} is a continuous surjection map then `f^(-1) (0) ∩ f^(-1) (1)` is:
Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0
The function f(x) = 5x – 3 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
For what value of `k` the following function is continuous at the indicated point
`f(x) = {{:(kx + 1",", if x ≤ pi),(cos x",", if x > pi):}` at = `pi`
Find the values of `a` and ` b` such that the function by:
`f(x) = {{:(5",", if x ≤ 2),(ax + b",", if 2 < x < 10),(21",", if x ≥ 10):}`
is a continuous function.
