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Question

The number of different possible values for the sum x+y+z, where x,y,z are real numbers such that x4+4y4+16z4+64=32xyz is


A
1
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B
2
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C
4
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D
8
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Solution

The correct option is C 4
Given :
x4+4y4+16z4+64=32xyz....... (i)
From the given equation we can say that 32xyz>0  .....(ii)
Now applying A.MG.M for the terms,
x4,4y4,16z4,64
x4+4y4+16z4+644(x4.4y4.16z4.64)14
using equation (i)
8xyz8xyz
A.M=G.M this condition holds when all the terms are equal,
x4=4y4=16z4=64x=±22y=±2z=±2
From equation (ii) 
Case 1 : when all three are positive 
1 way in which all are positive
Case 2 : when any 2 are negative 
(x,y) 
(z,y) 
(x,z)
so 3 ways 
Total possible combination of (x,y,z) is 4 
Hence number of different possible values for the sum = 4.

Mathematics

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