Advertisements
Advertisements
Question
\[f\left( x \right) = \frac{\left( 27 - 2x \right)^{1/3} - 3}{9 - 3 \left( 243 + 5x \right)^{1/5}}\left( x \neq 0 \right)\] is continuous, is given by
Options
\[\frac{2}{3}\]
6
2
4
Advertisements
Solution
2
For f(x) to be continuous at x = 0, we must have
\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - 3}{9 - 3 \left( 243 + 5x \right)^\frac{1}{5}}\]
\[ \Rightarrow f\left( 0 \right) = \lim_{x \to 0} \frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{3\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}\]
\[ = \frac{1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( {243}^\frac{1}{5} - \left( 243 + 5x \right)^\frac{1}{5} \right)}{x}}\]
\[ = \frac{- 1}{3} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{- 2x}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{5x}}\]
\[ = \frac{2}{15} \lim_{x \to 0} \frac{\frac{\left( 27 - 2x \right)^\frac{1}{3} - {27}^\frac{1}{3}}{27 - 2x - 27}}{\frac{\left( \left( 243 + 5x \right)^\frac{1}{5} - {243}^\frac{1}{5} \right)}{243 + 5x - 243}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times {27}^\frac{- 2}{3}}{\frac{1}{5} \times {243}^\frac{- 4}{5}}\]
\[ = \frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{{27}^\frac{2}{3}}}{\frac{1}{5} \times \frac{1}{{243}^\frac{4}{5}}}\]
\[ = 2\]
APPEARS IN
RELATED QUESTIONS
For what value of λ is the function defined by f(x) = `{(λ(x^2 - 2x)", if" x <= 0),(4x+ 1", if" x > 0):}` continuous at x = 0? What about continuity at x = 1?
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
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
\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0 .\]
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.
Discuss the continuity of the function
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.
If \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}\]
for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].
Show that f (x) = cos x2 is a continuous function.
What happens to a function f (x) at x = a, if
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 (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to
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 =
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
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 \[f\left( x \right) = x \sin\frac{1}{x}, x \neq 0,\]then the value of the function at x = 0, so that the function is continuous at x = 0, is
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
Let f (x) = |cos x|. Then,
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
If f(x) = 2x and g(x) = `x^2/2 + 1`, then which of the following can be a discontinuous function ______.
`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
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
The function f(x) = 5x – 3 is continuous at x =
What is the values of' 'k' so that the function 'f' is continuous at the indicated point
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 ______.
The function f(x) = x |x| is ______.
Discuss the continuity of the following function:
f(x) = sin x – cos x
