315 - Mathematics Questions Answers
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limit n tends to ∞
then
[³√(n²-n³) + n ] equals
We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]
Use the formula a3+b3 = (a+b)(a2+b2-ab) in the format
(a+b) = (a3+b3)/(a2+b2-ab)
here a = ³√(n²-n³) and b = n
on solving we get 1/(1+1+1) = 1/3
if f(X)=xlxl then find f-1(x) can you please explain me the meaning and use of sgn
f(x) = x|x| => f(x) = -x2 for negative real values of x and f(x) = x2 for positive real values of x
so f-1(x) = -√|x| for negative real values and = +√|x| for positive real values of x
The number of real negative terms in the binomial expansion of (1+ix)(4n-2), n € N and x > 0 is?
Total number of terms in the expansion = 4n-2+1 = 4n-1
for a negative real number i2 is compulsory so associated terms are 3rd, 7th, 11th ..........
Total number of terms = n
which of the following is differntiable at x=0 ?
cos(lxl)+lxl
sin(lxl)-lxl
sin(|x|)+|x| is differentiable at x = 0
LHD of first = {cos|0+h|+|0+h|-cos|0|-|0|}/0+h-0 = (cosh+h-1)/h = (1-2sin2h+h-1)/h = 1 (on taking limits)
now check RHD, its value will be -1
similarly in second both values are 0 so it is differentiable
Consider a branch of the hyperbola x²-2y²-2√2x-4√2y-6=0, with vertex at the point A. Let B be one of the end points of its latus rectum .If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is ??????
x²-2y²-2√2x-4√2y-6=0
Arrange this equation in form of standard hyperbola as
(x-√2)2/4 - (y+√2)2/2 = 1
so X = x-√2, Y = y+√2
vertex coordinate : X = 0 and Y = 0 so x = √2, y = -√2
similarly focus : X = ae, Y = 0, here a = 2, b = √2 and b2 = a2(1-e2)
and end point of latus rectum : X = ae, Y = b2/a
solve the area and get the answer