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Q.

 Let a1a2a3a4 be real numbers such that a1a2a3a4 = 0 and a21+a22+a23+a24 = 1. Then, the smallest possible value of the expression (a1- a2)2 +  ( a2- a3)2(a3- a4)2+(a4- a1)2 is:

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answer is 3.

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Detailed Solution

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Let's find the smallest possible value of the expression (a1- a2)2 +  ( a2- a3)2(a3- a4)2+(a4- a1)2

Given the conditions,  a1a2a3a4 = 0 and a21+a22+a23+a24 = 1.

First, we can simplify the expression:

(a1- a2)2 +  ( a2- a3)2(a3- a4)2+(a4- a1)2 

Expanding the squares:

a12+a22-2a1a2+a22+a32-2a2a3+a32+a42-2a3a4+a42+a12-2a4a1

Simplifying:

2a12+a22+a32+a42-2a1a2+a2a3+a3a4+a4a1   ....(1)

Since we know a12+a22+a32+a42=1 and a1+a2+a3+a4=0, we can rewrite the expression as:

2(1)-2a1a2+a2a3+a3a4+a4a1

Now, let's try to express a1a2+a2a3+a3a4+a4a1 in terms of the given conditions.

Squaring the sum a1+a2+a3+a4=0, we get:

a1+a2+a3+a42=a12+a22+a32+a42+2a1a2+a1a3+a1a4+a2a3+a2a4+a3a4)

Substituting the known values:

0=1+2a1a2+a1a3+a1a4+a2a3+a2a4+a3a4

Rearranging the terms:

a1a2+a1a3+a1a4+a2a3+a2a4+a3a4=-12

Now, we can substitute this value back into the equation (1):

2(1)-2a1a2+a2a3+a3a4+a4a1=2-2-12=3

So, the smallest possible value of the expression is 3.

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