45 - Algebra Questions Answers
The maximum number of real roots of the equation 2x88 + 3x87 − 13x2 + 5x + 9 = 0 is
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
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
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