166 - Questions Answers

(1+λ)ⁿ = C₀ + C₁x + ..................................C_nxⁿ

similarly for (1+μ)ⁿ and 

(λ+μ)ⁿ = C₀λⁿ + C₁λ^n-1μ + ........................ + C_nμⁿ

On multiplying these 3, we will get the coefficient of  λⁿμⁿ = ∑C_n³      (Here you should remember that C₀ = C_n, C₁ = C_n-1 and etc)

For getting the value of  ∑C_n³ :

multiply the expansions of (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ and calculate the coefficient of x⁰, we get ∑C_n³

and if we multiply (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ, then we get (-1)ⁿ(1-x²)²ⁿ/x²ⁿ so coefficient of x⁰ in this expression will be (-1)^n 2nC_n (-1)ⁿ  = ²ⁿC_n

Let x = (5√5 + 11)²ⁿ⁺¹ and y = (5√5 - 11)²ⁿ⁺¹

so x*y = 4²ⁿ⁺¹ = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)²ⁿ⁺¹ = (0.5)²ⁿ⁺¹  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even

let z = r(cosθ+i sinθ) and it is given that z¹⁵ = 1  =>  r¹⁵(cosθ+i sinθ)¹⁵ = 1

or r¹⁵(cos15θ+i sin15θ) = 1+0i                   (Here we use Demoivre's theorem)

on comparison we get sin 15θ = 0 => 15θ = nπ          here n is an integer

for n = -7 to +7 we will get 15 answers for which θ will lie between -π/2 to +π/2

max of sin and cos are 1 so max of left is 2

but x² + 1/x² ≥ 2x(1/x) = 2

so only solution will be obtained for 2cos²(x/2)sin²(x) = 2

so (1+cosx)(1-cos²x) = 2 which is impossible,so i think no solution

I think the last term would be cos(14π/15) 

We know that cos(14π/15) = -cos(π/15)

so     16cos(2π/15)cos(4π/15)cos(8π/15)cos(14π/15)

   = -16cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

   = -[8/sin(π/15)]2sin(π/15)cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

  =  -[8/sin(π/15)]sin(2π/15)cos(2π/15)cos(4π/15)cos(8π/15)   

take the similar steps                                          

2x⁸⁸ + 3x⁸⁷ − 13x² + 5x + 9 = 0

implies that 2x⁸⁸ + 3x⁸⁷ = 13x² - 5x - 9

or x⁸⁷(2x+3) = 13x² - 5x - 9

now plot the left and right graphs for getting the intersecting points

Hint: Parabola will face in upward direction as a is positive, take x=0 we get -9 so 1 root will be positive and 1 negative,

x⁸⁷(2x+3) will be 0 for x = -3/2 and x = 0 besides it consider some values as x = 1 and -1

You will get one root between 0 and -1 

now check both the graph at x= -2, you will get the idea about the second intersection

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^1/3ω, γ = -k^1/3ω²

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 = (k^1/3)³(-1) so cube root of p will be α = -k^1/3, β = -k^1/3ω, γ = -k^1/3ω² , here -1, -ω, -ω² are the cube roots of -1

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

x(-k^1/3)+y(-k^1/3ω)+z(-k^1/3ω²)/x(-k^1/3ω)+y(-k^1/3ω²)+z(-k^1/3)

=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

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

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)x^r

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)x^r

so required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132

let y = lim n->∞ (xⁿ/n!)

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

or y = 0