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Mathematics Question on Probability

If xx and yy are two real numbers such that x+y=1x + y = 1 and x3+y3=4,x^3 + y^3 = 4, then x5+y5x^5 + y^5 is

A

10

B

9

C

12

D

11

Answer

11

Explanation

Solution

We have, x+y=1⇒(x+y)2=1x + y = 1 \Rightarrow (x + y)2 = 1
⇒x2+y2+2xy=1\Rightarrow x^2 + y^2 + 2xy = 1 ... (i)
Also, x3+y3=4⇒(x+y)(x2−xy+y2)=4x^3 + y^3 = 4 \Rightarrow (x + y)(x^2 - xy + y^2) = 4
x2−xy+y2=4x^2 - xy + y^2 = 4 ... (ii)
Subtracting (ii) from (i), we get
x2+y2+2xy−x2+xy−y2=1−4x^2 + y^2 + 2xy - x^2 + xy - y^2 = 1 - 4
⇒3xy=−3⇒xy=−1\Rightarrow 3xy = -3 \Rightarrow xy = -1
(x+y)2(x3+y3)=(x2+y2+2xy)(x3+y3)(x + y)^2 (x^3 + y^3) = (x^2 + y^2 + 2xy)(x^3 + y^3)
(1).(4)=x5+x2y3+y2x3+y5+2.x4y+2xy4(1).(4) = x^5 + x^2y^3 + y^2x^3 + y^5 + 2.x^4y + 2xy^4
x5+y5=4−x2y3−y2x3−2x4y−2xy4x^5 + y^5 = 4 - x^2y^3 - y^2x^3 - 2x^4y - 2xy^4
=4−x2y2[y+x]−2xy[x3+y3]= 4- x^2y^2 [y + x] - 2xy[x^3 + y^3]
=4−x2y2−2xy(4)=4−(−1)2−2(−1)(4)= 4- x^2y^2- 2xy (4) = 4 - (-1)2- 2(-1)(4)
⇒x5+y5=4−1+8=11.\Rightarrow x^5+y^5=4-1+8=11.