Let a,b,c,d∈R+ and 256abcd≥(a+b+c+d)4 and 3a+b+2c+5d=11 then a3+b+c2+5d is
A
15
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B
8
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C
11
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D
20
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Solution
The correct option is B8 Given: 256abcd≥(a+b+c+d)4⇒(abcd)14≥(a+b+c+d)4 We know that A.M≥G.M (a+b+c+d)4≥(abcd)14 So both expression satisfies only when A.M=G.M ∴a=b=c=d We know that 3a+b+2c+5d=11⇒a=1=b=c=d Hence the value of the expression a3+b+c2+5d=8