Q.

 Let a1, a2, a3, a4 be real numbers such that a1+ a2+ a3+ a4 = 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 lies in the interval


see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

(0, 1.5)

b

(1.5, 2.5)

c

(2.5, 3)

d

(3, 3.5) 

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given that, a1+ a2+ a3+ a4 = 0 and a21+a22+a23+a24 = 1.
It is only possible when, a1=a2=12 and a3= a4=-12
Then, the given term becomes, (a1- a2)2 +  ( a2- a3)2+ (a3- a4)2+(a4- a1)2 =(12-12)2+(12+12)2+(-12+12)2+(-12-12)2
=0+1+0+1
=2
Hence the value lies between (1.5, 2.5)
Hence option (2) is correct.
 
Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
 Let a1, a2, a3, a4 be real numbers such that a1+ a2+ a3+ a4 = 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 lies in the interval