Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given:
\[ \Rightarrow \lim_{x \to 0} \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{1 - \sin^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2 \sin^2 x}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2\left( \sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{\left( \sqrt{x^2 + 1} - 1 \right)\left( \sqrt{x^2 + 1} + 1 \right)} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2\left( \ sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{x^2} = k\]
\[ \Rightarrow - 2 \lim_{x \to 0} \frac{\left( \ sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{x^2} = k\]
\[ \Rightarrow - 2 \lim_{x \to 0} \left( \frac{\ sinx}{x} \right)^2 \lim_{x \to 0} \left( \sqrt{x^2 + 1} + 1 \right) = k\]
\[ \Rightarrow - 2 \times 1 \times \left( 1 + 1 \right) = k\]
\[ \Rightarrow k = - 4\]
APPEARS IN
RELATED QUESTIONS
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.
Show that the function defined by f(x) = cos (x2) is a continuous function.
Find the values of a so that the function
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2kx}{x^2}, \text{ if } & x \neq 0 \\ 8 , \text{ if } & x = 0\end{cases}\] at x = 0
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:
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: \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if } x \neq 0 \\ 3k , & \text{ if } x = 0\end{cases}\]
The function f(x) is defined as follows:
If f is continuous on [0, 8], find the values of a and b.
Show that f (x) = | cos x | is a continuous 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.
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 =
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 function \[f\left( x \right) = \begin{cases}1 , & \left| x \right| \geq 1 & \\ \frac{1}{n^2} , & \frac{1}{n} < \left| x \right| & < \frac{1}{n - 1}, n = 2, 3, . . . \\ 0 , & x = 0 &\end{cases}\]
\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by
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 =
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 \[f\left( x \right) = \begin{cases}\frac{1 - \cos 10x}{x^2} , & x < 0 \\ a , & x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + \sqrt{x}} - 25}, & x > 0\end{cases}\] then the value of a so that f (x) may be continuous at x = 0, is
If \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]then the value of the function at x = 0, so that the function is continuous at x = 0, is
Find the values of a and b so that the function
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
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
A real value of x satisfies `((3 - 4ix)/(3 + 4ix))` = α – iβ (α, β ∈ R), if α2 + β2 is equal to
What is the values of' 'k' so that the function 'f' is continuous at the indicated point
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
The value of ‘k’ for which the function f(x) = `{{:((1 - cos4x)/(8x^2)",", if x ≠ 0),(k",", if x = 0):}` is continuous at x = 0 is ______.
