# 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

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

According to the given condition p is a negative number

consider p = -k so cube root of p will be α = -k^{1/3}, β = -k^{1/3}ω, γ = -k^{1/3}ω^{2}

On applying given values you will get the answer as ^{ }ω or ω^{2}

you should use ω^{3} = 1

^{ }

**Solution by Joshi sir**

According to the given condition p is a negative number

consider p = -k = (k^{1/3})^{3}(-1) so cube root of p will be α = -k^{1/3}, β = -k^{1/3}ω, γ = -k^{1/3}ω^{2 }, here -1, -ω, -ω^{2} are the cube roots of -1

On applying given values on the given expression, you will get

x(-k^{1/3})+y(-k^{1/3}ω)+z(-k^{1/3}ω^{2})/x(-k^{1/3}ω)+y(-k^{1/3}ω^{2})+z(-k^{1/3})

=x+yω+zω^{2}/xω+yω^{2}+z

first multiply ω in numerator and denominator we get answer as 1/ω = ω^{2}

second interchange the value of α, β, γ for getting answer as ω

you should use ω^{3} = 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**

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**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 2*x* + *y* + *z* = 21 is

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

You can solve this question by the following method

possile algebric expression for the given equation is

(x^{0}+x^{2}+x^{4}+........+x^{2}^{0})(x^{0}+x^{1}+x^{2}+..................+x^{21})(x^{0}+x^{1}+x^{2}+x^{3}+....................x^{21}) (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 (x^{0}+x^{2}+x^{4}+...........+x^{2}^{0})(1-x^{22})^{2}(1-x)^{-2}

now we have to calculate coefficient of x^{21 }in this expression so (1-x^{22})^{2} can be omitted

general term of (1-x)^{-2 }is (r+1)x^{r}

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

(x^{0}+x^{2}+x^{4}+........+x^{2}^{0})(x^{0}+x^{1}+x^{2}+..................+x^{21})(x^{0}+x^{1}+x^{2}+x^{3}+....................x^{21}) (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 (x^{0}+x^{2}+x^{4}+...........+x^{2}^{0})(1-x^{22})^{2}(1-x)^{-2}

now we have to calculate coefficient of x^{21 }in this expression so (1-x^{22})^{2} can be omitted

general term of (1-x)^{-2 }is (r+1)x^{r}

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**

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**Joshi sir comment**

let y = lim n->∞ (x^{n}/n!)

or y = x lim n->∞ (1/n) lim n->∞ x^{n-1}/(n-1)!

or y = 0

limit n tends to ∞

x{ [tan‾¹ (x+1/x+4)] - (π/4)}

**Asked By: AMIT DAS**

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**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

**Asked By: AMIT DAS**

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**Joshi sir comment**

We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]

Use the formula a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab) in the format

(a+b) = (a^{3}+b^{3})/(a^{2}+b^{2}-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**

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**Joshi sir comment**

f(x) = x|x| => f(x) = -x^{2} for negative real values of x and f(x) = x^{2} 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 **

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**N and x > 0 is?**

**Asked By: KAMAL**

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**Joshi sir comment**

Total number of terms in the expansion = 4n-2+1 = 4n-1

for a negative real number i^{2} 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**

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**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 = (1-2sin^{2}h+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**

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**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 b^{2} = a^{2}(1-e^{2})

and end point of latus rectum : X = ae, Y = b^{2}/a

solve the area and get the answer