Advertisements
Advertisements
प्रश्न
The value of f (0), so that the function
विकल्प
a3/2
a1/2
−a1/2
−a3/2
उत्तर
\[- a^\frac{1}{2}\]
Given:
\[\Rightarrow f\left( x \right) = \frac{\left( \sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( a^2 - ax + x^2 - \left( a^2 + ax + x^2 \right) \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( a + x - a + x \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( 2x \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{- a\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
If \[f\left( x \right)\] is continuous for all x, then it will be continuous at x = 0 as well.
So, if \[f\left( x \right)\] is continuous at x = 0, then
\[ \Rightarrow \left[ \frac{- 2a\left( \sqrt{a} \right)}{\left( \sqrt{a^2} + \sqrt{a^2} \right)} \right] = f\left( 0 \right)\]
\[ \Rightarrow \left[ \frac{- 2a\left( \sqrt{a} \right)}{\left( a + a \right)} \right] = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = - \sqrt{a}\]
APPEARS IN
संबंधित प्रश्न
Determine the value of 'k' for which the following function is continuous at x = 3
`f(x) = {(((x + 3)^2 - 36)/(x - 3), x != 3), (k, x = 3):}`
Examine the following function for continuity:
`f (x)1/(x - 5), x != 5`
Examine the following function for continuity:
`f(x) = (x^2 - 25)/(x + 5), x != -5`
Discuss the continuity of the function f, where f is defined by `f(x) = {(2x , ","if x < 0),(0, "," if 0 <= x <= 1),(4x, "," if x > 1):}`
Discuss the continuity of the following functions at the indicated point(s): (iv) \[f\left( x \right) = \left\{ \begin{array}{l}\frac{e^x - 1}{\log(1 + 2x)}, if & x \neq a \\ 7 , if & x = 0\end{array}at x = 0 \right.\]
Discuss the continuity of the following functions at the indicated point(s):
Discuss the continuity of the following functions at the indicated point(s):
Discuss the continuity of the function f(x) at the point x = 0, where \[f\left( x \right) = \begin{cases}x, x > 0 \\ 1, x = 0 \\ - x, x < 0\end{cases}\]
Discuss the continuity of \[f\left( x \right) = \begin{cases}2x - 1 & , x < 0 \\ 2x + 1 & , x \geq 0\end{cases} at x = 0\]
Determine the values of a, b, c for which the function f(x) = `{((sin(a + 1)x + sin x)/x, "for" x < 0),(x, "for" x = 0),((sqrt(x + bx^2) - sqrtx)/(bx^(3"/"2)), "for" x > 0):}` is continuous at x = 0.
If \[f\left( x \right) = \begin{cases}\frac{1 - \cos kx}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\text{is continuous at x} = 0, \text{ find } k .\]
If \[f\left( x \right) = \begin{cases}\frac{2^{x + 2} - 16}{4^x - 16}, \text{ if } & x \neq 2 \\ k , \text{ if } & x = 2\end{cases}\] is continuous at x = 2, find k.
Find the value of k for which \[f\left( x \right) = \begin{cases}\frac{1 - \cos 4x}{8 x^2}, \text{ when} & x \neq 0 \\ k ,\text{ when } & x = 0\end{cases}\] 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; \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5
Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = | x | + | x − 1 | at x = 0, 1.
Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if } x < \frac{\pi}{2} \\ a , & \text{ if } x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if } x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.
Find the values of a and b so that the function f(x) defined by \[f\left( x \right) = \begin{cases}x + a\sqrt{2}\sin x , & \text{ if }0 \leq x < \pi/4 \\ 2x \cot x + b , & \text{ if } \pi/4 \leq x < \pi/2 \\ a \cos 2x - b \sin x, & \text{ if } \pi/2 \leq x \leq \pi\end{cases}\]becomes continuous on [0, π].
Find all the points of discontinuity of f defined by f (x) = | x |− | x + 1 |.
Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.
Show that the function
\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.
Discuss the continuity and differentiability of
Write the points where f (x) = |loge x| is not differentiable.
If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text { is }\]
Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .
Examine the continuity of f(x)=`x^2-x+9 "for" x<=3`
=`4x+3 "for" x>3, "at" x=3`
Find the value of k for which the function f (x ) = \[\binom{\frac{x^2 + 3x - 10}{x - 2}, x \neq 2}{ k , x^2 }\] is continuous at x = 2 .
Find the value of 'k' if the function
f(x) = `(tan 7x)/(2x)`, for x ≠ 0.
= k for x = 0.
is continuous at x = 0.
Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if" x ≠ 0),(0",", "if" x = 0):}` is discontinuous at x = 0.
The number of points at which the function f(x) = `1/(x - [x])` is not continuous is ______.
For continuity, at x = a, each of `lim_(x -> "a"^+) "f"(x)` and `lim_(x -> "a"^-) "f"(x)` is equal to f(a).
Examine the continuity of the function f(x) = x3 + 2x2 – 1 at x = 1
f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if" x ≠ 2),(5",", "if" x = 2):}` at x = 2
f(x) = `{{:(|x - "a"| sin 1/(x - "a")",", "if" x ≠ 0),(0",", "if" x = "a"):}` at x = a
f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
f(x) = `{{:((1 - cos "k"x)/(xsinx)",", "if" x ≠ 0),(1/2",", "if" x = 0):}` at x = 0
Prove that the function f defined by
f(x) = `{{:(x/(|x| + 2x^2)",", x ≠ 0),("k", x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.
Find the values of a and b such that the function f defined by
f(x) = `{{:((x - 4)/(|x - 4|) + "a"",", "if" x < 4),("a" + "b"",", "if" x = 4),((x - 4)/(|x - 4|) + "b"",", "if" x > 4):}`
is a continuous function at x = 4.
If f is continuous on its domain D, then |f| is also continuous on D.
