Latest Elitmus Aptitude Question SOLUTION: a^2 + b^2 = 1/4 where a and b are integral values what would be the value of a and b for ab to be minimum? A) 1/sqrt8 B) -1/sqrt8 C) -1/8 D) 0

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(#M40035328) ELITMUS Numerical Ability question Quadratic equation Keep an EYE

a^2 + b^2 = 1/4
where a and b are integral values what would be the value of a and b for ab to be minimum?
A) 1/sqrt8
B) -1/sqrt8
C) -1/8
D) 0

Asked In Elitmus jitendra kumar (10 years ago)

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RAKESH  (10 years ago)

2ab =(a+b)^2 - (a^2+b^2)
=> ab = 1/2*(a+b)^2 - 1/2*(a^2+b^2)
=> ab = 1/2*(a+b)^2 - 1/8

as, 1/2*(a+b)^2 > 0 for any a,b
for ab to be minm 1/2*(a+b)^2 = 0 => a+b=0 => b=-a
& minm value of ab = -1/8
a^2 + b^2 = 1/4
=> 2*a^2 = 1/4
=> a^2 = 1/8
=> a = 1/sqrt8 , -1/sqrt8

however, a^2 + b^2 = 1/4 is not possible for any integral values of a & b.

minm value of ab = -1/8 for (a,b)= (1/sqrt8 , -1/sqrt8) or (-1/sqrt8 , 1/sqrt8)

Is this solution Helpfull? Yes (16) | No (2)  

sudhanshu kumar  (10 years ago)

a^2+b^2=1/4
we can write above as
=> (a+b)^2-2ab=1/4
=>2ab=(a+b)^2-1/4
=>ab= [(a+b)^2-1/4]/2
now,
ab is minimum when a+b=0 =>b= -a
so ab= -1/8
=> a(-a)=-1/8
=>-a^2=-1/8
=>a^2=1/8
=>a= +- 1/sqrt8
but in this problem a= 1/sqrt8
because in question it was mention that b

Is this solution Helpfull? Yes (9) | No   

Raj Kumar Ray  (10 years ago)

c) -1/8
here ab=1/2*(a+b)^2-1/8
so the 1st term i.e. 1/2*(a+b)^2 can never be negative so minimum value of this should be 0 for this. so min value for ab is -1/8
But the fact is a and b cannot be integers bcoz then a^2+b^2 can never be eaqual to 1/4

Is this solution Helpfull? Yes (1) | No   

nandini keshari  (10 years ago)

A)1/sqrt(8)

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adarsh  (10 years ago)

1/sqrt8 is the answer

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murthy  (10 years ago)

C is the correct answer

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jitendra kumar  (10 years ago)

for this que if we go through the options
then, for A>
b^2 = 1/4 - 1/8
b = 1/ sqrt8 so, ab = 1/8
for B>
b^2 = 1/4- 1/8
b = 1/sqrt8 so, ab = -1/8
for C>
b^2 = 1/4 - 1/64
b = sqrt(15/64) = sqrt(15)/8
so, ab = -sqrt(15)/64
for D>
ab = 0
so from all the options
A > D > B > C
so for ab to be minimum ans = C> -1/8

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sudhanshu kumar  (10 years ago)

now continue from my above answer

ant it was mention in question that b

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abhishek  (9 years ago)

for two nos x and y ..A.M(x,y)>= G.M.(x,y)
put x= a^2,y =b ^2..
(a^2+b^2)/2 >= sqrt(a^2*b^2);
1/8 >=sqrt(a^2*b^2);
sqrt(1/8)>=|ab|;
hence ab belongs to {-sqrt(-1/8),sqrt(1/8)};

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jitendra  (8 years ago)

(a+b)^2=a^2+b^2+2ab
(a+b)^2>=0
a^2+b^2+2ab>=0
1/4+2ab>=0
ab>= - 1/8

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