# 320 - Mathematics Questions Answers

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Submit By: MANISH SIR 6 year ago
is this topic helpfull: 26 7

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

Asked By: HIMANSHU MITTAL 8 year ago
is this question helpfull: 5 0 read solutions ( 3 ) | submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Solution by manish sir

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 = -k1/3ω, 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

Asked By: HIMANSHU MITTAL 8 year ago
is this question helpfull: 6 1 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Asked By: HIMANSHU MITTAL 8 year ago
is this question helpfull: 8 1 read solutions ( 1 ) | submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Solution by manish sir

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

Asked By: AMIT DAS 8 year ago
is this question helpfull: 5 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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)}

Asked By: AMIT DAS 8 year ago
is this question helpfull: 6 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Solution by manish sir

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

Asked By: AMIT DAS 8 year ago
is this question helpfull: 5 1 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Asked By: NIKHIL VARSHNEY 8 year ago
is this question helpfull: 18 2 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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?

Asked By: KAMAL 8 year ago
is this question helpfull: 6 0 read solutions ( 1 ) | submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

Asked By: NIKHIL VARSHNEY 8 year ago
is this question helpfull: 11 3 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 ??????

Asked By: AMIT DAS 8 year ago
is this question helpfull: 6 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

## 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

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