Basic maths-calculus-Q18

+1 vote

Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true ? 

(A) The minimum possible value of a^2 + b^2 + c^2 + d^2 is 4m^2-2m+1

(B) The minimum possible value of a^2 + b^2 + c^2 + d^2 is 4m^2+2m+1

(C) The maximum possible value of a^2 + b^2 + c^2 + d^2 is 4m^2-2m+1

(D) The maximum possible value of a^2 + b^2 + c^2 + d^2 is 4m^2+2m+1

asked May 19 in Basic Maths by gbeditor (44,560 points)
reshown May 24 by gbeditor

1 Answer

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Best answer

m is a positive integer, so, lowest possible value of m=1.

Hence, in order to a+b+c+d = 4m+1, when m=1, at least one of a,b,c,d has to be greater than 1.

Take a=b=c=1; d=2

Hence, a^2+b^2+c^2+d^2 = 7

Which is equal to 4m^2 + 2m+1 , for other values of a,b,c,d, a^2+b^2+c^2+d^2  will be greater than 4m^2 + 2m+1

Hence, answer is B.

answered May 27 by deepak-gatebook (112,760 points)
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