Advertisements
Advertisements
प्रश्न
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) `{:(= (x^3 - 8)/(x^2 - 4)",", "for" x > 2),(= 3",", "for" x = 2),(= ("e"^(3(x - 2)^2 - 1))/(2(x - 2)^2) ",", "for" x < 2):}`
Advertisements
उत्तर
f(2) = 3 ...(Given)
`lim_(x -> 2^-) "f"(x) = lim_(x -> 2^-) (x^3 - 8)/(x^2 - 4)`
= `lim_(x -> 2^-) (x^3 - 2^3)/(x^2 - 2^2)`
= `lim_(x -> 2^-) ((x - 2) (x^2 + 2x + 4))/((x - 2)(x + 2))`
= `lim_(x -> 2^-) (x^2 + 2x + 4)/(x + 2)`
= `(lim_(x -> 2^-) (x^2 + 2x + 4))/(lim_(x -> 2^-) (x + 2))`
= `((2)^2 + 2(2) + 4)/(2 + 2)`
= `12/4`
= 3
`lim_(x -> 2^+) "f"(x) = lim_(x -> 2^+) ("e"^(3(x - 2)^2) - 1)/(2(x - 2)^2)`
Put x – 2 = h
∴ x = 2 + h
As x → 2, h → 0
∴ `lim_(x -> 2^+) "f"(x) = lim_("h" -> 0) ("e"^(3"h"^2) - 1)/(2"h"^2)`
= `1/2 lim_("h" -> 0) ("e"^(3"h"^2) - 1)/(3"h"^2) xx 3`
= `1/2 xx 1 xx 3 ...[(because "h" -> 0 therefore "h"^2 -> 0),(and lim_(x -> 0) ("e"^x - 1)/x = 1)]`
= `3/2`
∴ `lim_(x -> 2^-) "f"(x) ≠ lim_(x -> 2^+) "f"(x)`
∴ `lim_(x -> 2) "f"(x)` does not exist
∴ f(x) is discontinuous at x = 2
This discontinuity is irremovable.
APPEARS IN
संबंधित प्रश्न
Examine whether the function is continuous at the points indicated against them:
f(x) = x3 − 2x + 1, for x ≤ 2
= 3x − 2, for x > 2, at x = 2
Examine whether the function is continuous at the points indicated against them:
f(x) = `(x^2 + 18x - 19)/(x - 1)` for x ≠ 1
= 20 for x = 1, at x = 1
Examine the continuity of f(x) = x3 + 2x2 − x − 2 at x = − 2
Examine the continuity of `"f"(x) {:(= sin x",", "for" x ≤ pi/4), (= cos x",", "for" x > pi/4):}} "at" x = pi/4`
Examine whether the function is continuous at the points indicated against them:
f(x) `{:(= x^3 - 2x + 1",", "if" x ≤ 2),(= 3x - 2",", "if" x > 2):}}` at x = 2
Examine whether the function is continuous at the points indicated against them :
f(x) `{:( = (x^2 + 18x - 19)/(x - 1)",", "for" x ≠ 1),(= 20",", "for" x = 1):}}` at x = 1
Examine whether the function is continuous at the points indicated against them :
f(x) `{:(= x/(tan3x) + 2",", "for" x < 0),(= 7/3",", "for" x ≥ 0):}} "at" x = 0`
Find all the point of discontinuities of f(x) = [x] on the interval (− 3, 2).
Discuss the continuity of the function f(x) = |2x + 3|, at x = `(−3)/(2)`
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= (sqrt(x - 1) - (x - 1)^(1/3))/(x - 2)",", "for" x ≠ 2),(= 1/5",", "for" x = 2):}}`at x = 2
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= (x^3 - 8)/(sqrt(x + 2) - sqrt(3x - 2))",", "for" x ≠ 2),(= -24",", "for" x = 2):}}` at x = 2
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= ((27 - 2x)^(1/3) - 3)/(9 - 3(243 + 5x)^(1/5))",", "for" x ≠ 0),(= 2",", "for" x = 0):}}` at x = 0.
Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :
f(x) = `(x^2 - 1)/(x^3 + 1)` for x ≠ – 1
Discuss the continuity of the following function at the point indicated against them :
f(x) = `{:(=( sqrt(3) - tanx)/(pi - 3x)",", x ≠ pi/3),(= 3/4",", x = pi/3):}} "at" x = pi/3`
If f(x) = `(sqrt(2 + sin x) - sqrt(3))/(cos^2x), "for" x ≠ pi/2`, is continuous at x = `pi/2` then find `"f"(pi/2)`
Show that there is a root for the equation x3 − 3x = 0 between 1 and 2.
Select the correct answer from the given alternatives:
If f(x) = `(1 - sqrt(2) sinx)/(pi - 4x), "for" x ≠ pi/4` is continuous at x = `pi/4`, then `"f"(pi/4)` =
Select the correct answer from the given alternatives:
If f(x) = [x] for x ∈ (–1, 2) then f is discontinuous at
Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
f(x) `{:(= 2x^2 - 2x + 5",", "for" 0 ≤ x ≤ 2),(= (1 - 3x - x^2)/(1 - x) "," , "for" 2 < x < 4),(= (x^2 - 25)/(x - 5)",", "for" 4 ≤ x ≤ 7 and x ≠ 5),(= 7",", "for" x = 5):}`
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
f(x) `{:(= x^2 + 5x + 1"," , "for" 0 ≤ x ≤ 3),(= x^3 + x + 5"," , "for" 3 < x ≤ 6):}`
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
f(x) `{:(= (x^2 + x + 1)/(x + 1)"," , "for" x ∈ [0, 3)),(=(3x +4)/(x^2 - 5)"," , "for" x ∈ [3, 6]):}`
Discuss the continuity of the following function at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous:
f(x) = `((x + 3)(x^2 - 6x + 8))/(x^2 - x - 12)`
Find k if following function is continuous at the point indicated against them:
f(x) `{:(= ((5x - 8)/(8 - 3x))^(3/(2x - 4))",", "for" x ≠ 2),(= "k"",", "for" x = 2):}}` at x = 2
Solve using intermediate value theorem:
Show that x3 − 5x2 + 3x + 6 = 0 has at least two real root between x = 1 and x = 5
If f(x) = `{:{(tan^-1|x|; "when" x ≠ 0), (pi/4; "when" x = 0):}`, then ______
If f(x) = `[tan (pi/4 + x)]^(1/x)`, x ≠ 0 at
= k, x = 0 is continuous x = 0. Then k = ______.
If f(x) = `{(8-6x; 0<x≤2), (4x-12; 2<x≤3),(2x+10; 3<x≤6):}` then f(x) is ______
If function `f(x)={((x^2-9)/(x-3), ",when "xne3),(k, ",when "x =3):}` is continuous at x = 3, then the value of k will be ______.
If f(x) = `{{:(tanx/x + secx",", x ≠ 0),(2",", x = 0):}`, then ______.
Let f be the function defined by
f(x) = `{{:((x^2 - 1)/(x^2 - 2|x - 1| - 1)",", x ≠ 1),(1/2",", x = 1):}`
If f(x) = `{{:((x - 4)/(|x - 4|) + a",", "for" x < 4),(a + b",", "for" x = 4 "is continuous at" x = 4","),((x - 4)/(|x - 4|) + b",", "for" x > 4):}`
then ______.
If f(x) = `{{:((3 sin πx)/(5x),",", x ≠ 0),(2k,",", x = 0):}`
is continuous at x = 0, then the value of k is ______.
`lim_(x rightarrow 0) (e^(x^2) - cosx)/x^2` is equal to ______.
