Advertisements
Advertisements
प्रश्न
Find the value of k if f(x) is continuous at x = π/2, where \[f\left( x \right) = \begin{cases}\frac{k \cos x}{\pi - 2x}, & x \neq \pi/2 \\ 3 , & x = \pi/2\end{cases}\]
Advertisements
उत्तर
Given: \[f\left( x \right) = \binom{\frac{k\ cosx}{\pi - 2x}, x \neq \frac{\pi}{2}}{3, x = \frac{\pi}{2}}\]
If f(x) is continuous at x =
\[\frac{\pi}{2}\]then
\[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
\[\lim_{x \to \frac{\pi}{2}} \frac{k\ cosx}{\pi - 2x} = 3\]
Putting
\[\frac{\pi}{2} - x = h\]
\[\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} =\]
\[\lim_{h \to 0} \frac{k \cos \left( \frac{\pi}{2} - h \right)}{\pi - 2\left( \frac{\pi}{2} - h \right)}\]
From (1), we have
\[\lim_{h \to 0} \frac{k \cos \left( \frac{\pi}{2} - h \right)}{\pi - 2\left( \frac{\pi}{2} - h \right)} = 3\]
\[\Rightarrow \lim_{h \to 0} \frac{k \sin h}{2h} = 3\]
\[\Rightarrow \lim_{h \to 0} \frac{k \sin h}{h} = 6\]
\[\Rightarrow k \lim_{h \to 0} \frac{\sin h}{h} = 6\]
\[\Rightarrow k \times 1 = 6\]
\[\Rightarrow k = 6\]
Hence, for
\[k = 6\] , f(x) is continuous at x =
\[\frac{\pi}{2}\]
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`
Find the values of a and b such that the function defined by f(x) = `{(5", if" x <= 2),(ax +b", if" 2 < x < 10),(21", if" x >= 10):}` is a continuous function.
Examine that sin |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.
Find the values of a so that the function
Let \[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}\] x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.
If \[f\left( x \right) = \begin{cases}\frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1}, & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, find k.
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.
Discuss the continuity of the function
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: \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if } x < 0 \\ \cos 2x , & \text{ if } x \geq 0\end{cases}\]
The function f(x) is defined as follows:
If f is continuous on [0, 8], find the values of a and b.
Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.
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 \[f\left( x \right) = \begin{cases}\frac{\sin^{- 1} x}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]is continuous at x = 0, write the value of 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 =
The function \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\]
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 (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 is defined by \[f\left( x \right) = x^2 - 4x + 7\] , show that \[f'\left( 5 \right) = 2f'\left( \frac{7}{2} \right)\]
If \[f\left( x \right) = a\left| \sin x \right| + b e^\left| x \right| + c \left| x \right|^3\]
The function f (x) = x − [x], where [⋅] denotes the greatest integer function is
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(x) = 2x and g(x) = `x^2/2 + 1`, then which of the following can be a discontinuous function ______.
`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 ____________.
The point(s), at which the function f given by f(x) = `{("x"/|"x"|"," "x" < 0),(-1"," "x" ≥ 0):}` is continuous, is/are:
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
Let f(x) = `{{:(5^(1/x), x < 0),(lambda[x], x ≥ 0):}` and λ ∈ R, then at x = 0
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
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.
Discuss the continuity of the following function:
f(x) = sin x + cos x
