Advertisements
Advertisements
प्रश्न
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}4 , & \text{ if } x \leq - 1 \\ a x^2 + b, & \text{ if } - 1 < x < 0 \\ \cos x, &\text{ if }x \geq 0\end{cases}\]
Advertisements
उत्तर
Given:
If \[f\left( x \right)\] is continuous at x = −1 and 0, then
\[\Rightarrow \lim_{h \to 0} f\left( - 1 - h \right) = \lim_{h \to 0} f\left( - 1 + h \right) \text{ and } \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right) \]
\[ \Rightarrow \lim_{h \to 0} \left( 4 \right) = \lim_{h \to 0} \left( a \left( - 1 + h \right)^2 + b \right) \text{ and } \lim_{h \to 0} \left( a \left( - h \right)^2 + b \right) = \lim_{h \to 0} \left( \cos h \right)\]
\[ \Rightarrow 4 = a + b \text{ and } b = 1\]
\[ \Rightarrow a = 3 \text{ and } b = 1\]
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.
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`
Show that the function defined by f(x) = |cos x| is a continuous function.
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
\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0 .\]
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
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:
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: \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}\]
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.
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 =
If \[f\left( x \right) = \begin{cases}\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] and f (x) is continuous at x = 0, then the value of k is
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
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
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
The value of a for which the function \[f\left( x \right) = \begin{cases}5x - 4 , & \text{ if } 0 < x \leq 1 \\ 4 x^2 + 3ax, & \text{ if } 1 < x < 2\end{cases}\] is continuous at every point of its domain, is
If \[f \left( x \right) = \sqrt{x^2 + 9}\] , write the value of
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
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
`lim_("x" -> 0) ("x cos x" - "log" (1 + "x"))/"x"^2` 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
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
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 =
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
The function f(x) = x |x| is ______.
