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) = `((3 - 8x)/(3 - 2x))^(1/x)`, for x ≠ 0
Advertisements
उत्तर
f(x) = `((3 - 8x)/(3 - 2x))^(1/x)`, for x ≠ 0
Here, f(0) is not defined.
Consider, `lim_(x -> 0) "f"(x) = lim_(x -> 0) ((3 - 8x)/(3 - 2x))^(1/x)`
= `lim_(x -> 0) [(3(1 - (8x)/3))/(3(1 - (2x)/3))]^(1/x)`
= `lim_(x -> 0) ((1 - (8x)/3)^(1/x))/(1 - (2x)/3)^(1/x)`
= `(lim_(x -> 0) [(1 - (8x)/3)^((-3)/(8x))]^((-8)/3))/(lim_(x -> 0) [(1 - (2x)/3)^((-3)/(2x))]^((-2)/3))`
= `("e"^((-8)/3))/("e"^((-2)/3)) ...[(because x -> 0"," (-8x)/3 -> 0"," (-2x)/3 -> 0),(and lim_(x -> 0) (1 + x)^(1/x) = "e")]`
= `"e"^((-6)/3)`
= e–2
∴ `lim_(x -> 0) "f"(x)` exists
But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
This discontinuity can be removed by defining f(0) = e–2
∴ f(x) can be redefined as
f(x) `{:(= ((3 - 8x)/(3 - 2x))^(1/x), ";" x ≠ 0),(= "e"^(-2), ";" x = 0):}`
APPEARS IN
संबंधित प्रश्न
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`
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
Identify discontinuities for the following function as either a jump or a removable discontinuity :
f(x) `{:(= x^2 - 3x - 2",", "for" x < -3),(= 3 + 8x",", "for" x > -3):}`
Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :
f(x) = `(3sin^2 x + 2cos x(1 - cos 2x))/(2(1 - cos^2x)`, for x ≠ 0
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)`
Select the correct answer from the given alternatives:
f(x) = `{:(= (2^(cotx) - 1)/(pi - 2x)",", "for" x ≠ pi/2),(= log sqrt(2)",", "for" x = pi/2):}`
Select the correct answer from the given alternatives:
If f(x) = `((sin2x)tan5x)/("e"^(2x) - 1)^2`, for x ≠ 0 is continuous at x = 0, then f(0) is
Select the correct answer from the given alternatives:
f(x) = `(x^2 - 7x + 10)/(x^2 + 2x - 8)`, for x ∈ [– 6, – 3]
Select the correct answer from the given alternatives:
If f(x) `{:(= "a"x^2 + "b"x + 1",", "for" |x −1| ≥ 3), (= 4x + 5",", "for" -2 < x < 4):}` is continuous everywhere then,
Select the correct answer from the given alternatives:
f(x) `{:(= (32^x - 8^x - 4^x + 1)/(4^x - 2^(x + 1) + 1)",", "for" x ≠ 0),(= "k""," , "for" x = 0):}` is continuous at x = 0, then value of ‘k’ is
Select the correct answer from the given alternatives:
If f(x) = `((4 + 5x)/(4 - 7x))^(4/x)`, for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
f(x) = [x + 1] for x ∈ [−2, 2)
Where [*] is greatest integer function.
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
f(x) `{:(= x^2 + x - 3,"," "for" x ∈ [ -5, -2)),(= x^2 - 5,"," "for" x ∈ (-2, 5]):}`
Find a and b if following function is continuous at the point or on the interval indicated against them:
f(x) `{:(= (4tanx + 5sinx)/("a"^x - 1)",", "for" x < 0),(= (9)/(log2)",", "for" x = 0),(= (11x + 7x*cosx)/("b"^x - 1)",", "for" x > 0):}`
Find a and b if following function is continuous at the point or on the interval indicated against them:
f(x) `{:(= "a"x^2 + "b"x + 1",", "for" |2x - 3| ≥ 2),(= 3x + 2",", "for" 1/2 < x < 5/2):}`
Solve using intermediate value theorem:
Show that 5x − 6x = 0 has a root in [1, 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 (pi/4 + x)]^(1/x)`, x ≠ 0 at
= k, x = 0 is continuous x = 0. Then k = ______.
Let f : [-1, 2] → [0, ∞] be a continuous function such that f(x) = f(1 - x) ∀ x ∈ [-1, 2].
Let R1 = `int_-1^2 xf(x) dx` and R2 be the area of the region bounded by y = f(x), x = -1, x = 2 and the X-axis. Then, ______
If f(x) = `{(8-6x; 0<x≤2), (4x-12; 2<x≤3),(2x+10; 3<x≤6):}` then f(x) is ______
If f(x) = `{{:(tanx/x + secx",", x ≠ 0),(2",", x = 0):}`, then ______.
If f(x) = `{{:((sin5x)/(x^2 + 2x)",", x ≠ 0),(k + 1/2",", x = 0):}` is continuous at x = 0, then the value of k is ______.
If f(x) = `{{:(x, "for" x ≤ 0),(0,
"for" x > 0):}`, then f(x) at x = 0 is ______.
For x > 0, `lim_(x rightarrow 0) ((sin x)^(1//x) + (1/x)^sinx)` is ______.
If f(x) = `{{:((3 sin πx)/(5x),",", x ≠ 0),(2k,",", x = 0):}`
is continuous at x = 0, then the value of k is ______.
If \[\mathrm{f}(x)= \begin{cases} \mathrm{m}x+1, & x\leqslant\frac{\pi}{2} \\ \\ \mathrm{sin}x+\mathrm{n}, & x>\frac{\pi}{2} & \end{cases}\], is continuous at \[x=\frac{\pi}{2},( \begin{array} {c}\mathrm{m,n\in\mathbb{Z}} \end{array})\] then
If \[f\] is defined on the closed interval \[[a,b]\], which condition establishes continuity at the right endpoint \[b\]?
Which condition makes a function discontinuous at \[x=a\]?
What is the graphical meaning of a function being continuous at a point?
For which values is the constant function \[f(x)=k\] continuous?
