320 - Mathematics Questions Answers
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Let α, β and γ be the cube roots of p (p < 0). For any real numbers x, y and z, the value of (xα+yβ+zγ) ⁄(xβ+yγ+zα) is
According to the given condition p is a negative number
consider p = -k so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω2
On applying given values you will get the answer as ω or ω2
you should use ω3 = 1
According to the given condition p is a negative number
consider p = -k = (k1/3)3(-1) so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω2 , here -1, -ω, -ω2 are the cube roots of -1
On applying given values on the given expression, you will get
x(-k1/3)+y(-k1/3ω)+z(-k1/3ω2)/x(-k1/3ω)+y(-k1/3ω2)+z(-k1/3)
=x+yω+zω2/xω+yω2+z
first multiply ω in numerator and denominator we get answer as 1/ω = ω2
second interchange the value of α, β, γ for getting answer as ω
you should use ω3 = 1
If A and B are two independent events such that and P(A'∩B)=2/7 and P(A∩B')=1/5, then P(A) is
Let p(A) = x and p(B) = y
we know that p(A∩B) + p(A∩B') = p(A)
and p(A∩B) + p(B∩A') = p(B)
and for independent event p(A∩B) = p(A)p(B)
so on applying all the given conditions relation etween x and y will be
xy + 1/5 = x
xy + 2/7 = y
solve now
Total number of non-negative integral solutions of 2x + y + z = 21 is
You can solve this question by the following method
possile algebric expression for the given equation is
(x0+x2+x4+........+x20)(x0+x1+x2+..................+x21)(x0+x1+x2+x3+....................x21) (i)
Here three brackets are for x , y and z, and powers are based on the range of least to greatest possile values of x, y, z
on solving eq (i) will become (x0+x2+x4+...........+x20)(1-x22)2(1-x)-2
now we have to calculate coefficient of x21 in this expression so (1-x22)2 can be omitted
general term of (1-x)-2 is (r+1)xr
so required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132
You can solve this question by the following method
possile algebric expression for the given equation is
(x0+x2+x4+........+x20)(x0+x1+x2+..................+x21)(x0+x1+x2+x3+....................x21) (i)
Here three brackets are for x , y and z, and powers are based on the range of least to greatest possile values of x, y, z
on solving eq (i) will become (x0+x2+x4+...........+x20)(1-x22)2(1-x)-2
now we have to calculate coefficient of x21 in this expression so (1-x22)2 can be omitted
general term of (1-x)-2 is (r+1)xr
so required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132
limit n tends to ∞
then
(x^n)/(n!) equals
let y = lim n->∞ (xn/n!)
or y = x lim n->∞ (1/n) lim n->∞ xn-1/(n-1)!
or y = 0
limit n tends to ∞
x{ [tan‾¹ (x+1/x+4)] - (π/4)}
I think n is printed by mistake, so considering it as x
the question will be limit x tends to ∞
x{ [tan-1 (x+1/x+4)] - (π/4)}
consider [tan-1 (x+1/x+4)] - (π/4) = θ
so [tan-1 (x+1/x+4)] = (π/4)+θ
so (x+1/x+4) = tan(π/4+θ)
or x = (-3-5tanθ)/2tanθ
so converted question will be limθ->0 (-3-5tanθ)θ/2tanθ
solve
I think n is printed by mistake, so considering it as x
the question will be limit x tends to ∞
x{ [tan-1 (x+1/x+4)] - (π/4)}
consider [tan-1 (x+1/x+4)] - (π/4) = θ
so [tan-1 (x+1/x+4)] = (π/4)+θ
so (x+1/x+4) = tan(π/4+θ)
or x = (-3-5tanθ)/2tanθ
so converted question will be limθ->0 (-3-5tanθ)θ/2tanθ
now we know that limθ->0 tanθ/θ = limθ->0 θ/tanθ = 1
so next line will be limθ->0 (-3-5tanθ)/2 = -3/2
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