Elitmus
Exam
Numerical Ability
Number System
if (1/a^2 )+ (1/b^2 )= 1/4. then what could be the minimum value of ab ? .(a,b are integers and a>b)
Read Solution (Total 13)
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- 1/a^2 + 1/b^2=1/4
=>(a^2+b^2)=(a.b)^2/4;
(a+b)^2=a^2+b^2+2.a.b
=(a.b)^2/4 +2.a.b
=ab(ab+8)/4
so either ,
ab=0 or ab=-8; - 9 years agoHelpfull: Yes(23) No(5)
- pushpak ur ans is wrong bcoz if u r taking b=0 i.e 1/0^2 which is impossible
- 9 years agoHelpfull: Yes(5) No(0)
- ans is 0 assume a=2 b=0 then ans is 1/4 and a*b =0
- 9 years agoHelpfull: Yes(4) No(13)
- ans is -64
a=8 and b=-8 because condition is a>b
substitute the values and get the answer is -64 - 9 years agoHelpfull: Yes(3) No(3)
- 1/a^2+1/b^2=1/4
=>(b^2+a^2)/(ab)^2=1/4
=>(ab)^2= 4(b^2+a^2)
=>ab= 2 sqrt(a^2+b^2)
Let a =1 and b=0 because a>b
ANS ab=2
- 9 years agoHelpfull: Yes(3) No(4)
- Sorry the above Given process is correct but a=2√2 and b=-2√2 so ab=-8 then it satisfies all above conditions because a>b
- 9 years agoHelpfull: Yes(3) No(0)
- Ans a.b= -8
a=2 sqrt2 and b= -2 sqrt2
1/(2 sqrt2)^2 + 1/(-2 sqrt2)^2= 1/8+1/8=2/8=1/4 - 9 years agoHelpfull: Yes(2) No(1)
- 1/8+1/8=1/4
so may be ab=8 - 9 years agoHelpfull: Yes(2) No(5)
- AM=>GM ans 8
- 9 years agoHelpfull: Yes(2) No(0)
- (b^2+a^2)/a^2*b^2=1/4
4(b^2+a^2)=a^2*b^2
4b^2+4a^2=a^2*b^2
√(4a^2+4b^2)=ab
so a should be greater than b
so consider minimum value a=-2√2,b=-2√2
then ab=8 The a,b can satisfy all above conditions
a and b cannot be 0 because 1/0 is undefined - 9 years agoHelpfull: Yes(0) No(2)
- put the value of a as 4 and b as 3
so we get the answer as ab=10(2*5)
- 9 years agoHelpfull: Yes(0) No(0)
- A to q, a and b are integers so ab must be integer...let a=8 and b=6 then, ab= sqrt root(a^2+b^2)/2,therefore ab is 5.
- 9 years agoHelpfull: Yes(0) No(0)
- Acc to q, a and b are integers so ab must be integer.
let a=8 and b=6
therefore ab=(sqrt(a^2+b^2))/2
put the the value of a and b in above equation so we get
the minimum value of ab is equal to 5. - 9 years agoHelpfull: Yes(0) No(0)
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