# 160 - Mathematics Questions Answers

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

Joshi sir comment

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 Joshi 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

Joshi sir comment

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

Joshi sir comment

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 Joshi 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

Joshi sir comment

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)}

Joshi sir comment

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 Joshi 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

Joshi sir comment

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

Joshi sir comment

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?

Joshi sir comment

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

Joshi sir comment

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 ??????

Joshi sir comment

## 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