Advertisements
Advertisements
प्रश्न
Find the values 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 " = pi`
उत्तर
`f(x) = {(kx +1, if x<= pi),(cos x, if x > pi):} " at x " = pi`
If f(x) is continuous at `x = pi` it implies:
`f(pi) = lim_(x -> pi^+) f(x) = lim_(x -> pi^-) f(x)`
`=> k (pi) + 1 = cos (pi) = k (pi) + 1`
`=> k (pi) + 1 = cos (pi)`
`=> k (pi) + 1 = - 1`
`=> k = (-2)/pi`
That is, for the quantity `k = (-2)/pi` this function is continuous at `x = pi`.
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 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 cosine, cosecant, secant and cotangent functions,
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
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.
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: \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}2 , & \text{ if } x \leq 3 \\ ax + b, & \text{ if } 3 < x < 5 \\ 9 , & \text{ if } x \geq 5\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}4 , & \text{ if } x \leq - 1 \\ a x^2 + b, & \text{ if } - 1 < x < 0 \\ \cos x, &\text{ if }x \geq 0\end{cases}\]
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 \[f\left( x \right) = \begin{cases}\frac{1 - \sin x}{\left( \pi - 2x \right)^2} . \frac{\log \sin x}{\log\left( 1 + \pi^2 - 4\pi x + 4 x^2 \right)}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]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
\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by
The function
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
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 \[f\left( x \right) = \begin{cases}\frac{\left| x + 2 \right|}{\tan^{- 1} \left( x + 2 \right)} & , x \neq - 2 \\ 2 & , x = - 2\end{cases}\] then f (x) is
If \[f\left( x \right) = a\left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3\]
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
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) = `"e"^|x|` is ______.
If f.g is continuous at x = a, then f and g are separately continuous at x = a.
`lim_("x"-> pi) (1 + "cos"^2 "x")/("x" - pi)^2` is equal to ____________.
`lim_("x" -> 0) ("x cos x" - "log" (1 + "x"))/"x"^2` is equal to ____________.
`lim_("x" -> 0) (1 - "cos" 4 "x")/"x"^2` is equal to ____________.
Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0
Discuss the continuity of the following function:
f(x) = sin x + cos x
Discuss the continuity of the following function:
f(x) = sin x – cos x