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13 Mar 2020, 03:36
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A
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Difficulty:
65%
(hard)
Question Stats:
64% (02:29) correct
36%
(02:15)
wrong
based on 244
sessions
History
Date
Time
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Not Attempted Yet
If \(F_{1} (a,b) = \frac{(a-b)}{(a+b)}\) and \(F_{2} (a,b) = \frac{1}{(a+b)^2 }\)then for positive integers (a, b) the maximum value of \( F^2_{1} (a,b) + F_{2} (a,b)\) is?
A) less than 1
B) 0
C) -1
D) occurs at a=b
E) cannot be estimated
Kudos for a correct solution!
A) less than 1
B) 0
C) -1
D) occurs at a=b
E) cannot be estimated
Kudos for a correct solution!
13 Mar 2020, 06:29
Kudos
Bookmarks
\( F^2_{1} (a,b) + F_{2} (a,b)\)
= \(\frac{(a-b)^2}{(a+b)^2} + \frac{1}{(a+b)^2 }\)
= \(\frac{(a-b)^2+1}{(a+b)^2} \)
= \(\frac{(a^2+b^2-2ab +1)}{(a+b)^2} \)
= \(\frac{(a^2+b^2+2ab -4ab +1)}{(a+b)^2} \)
= \(\frac{(a+b)^2}{(a+b)^2} - \frac{(4ab-1)}{(a+b)^2}\)
= \(1 - \frac{(4ab-1)}{(a+b)^2}\)
= 1 - (positive real number).......... {as a and b are positive integers, 4ab-1 ≥ 4*1*1 -1=3}
\( F^2_{1} (a,b) + F_{2} (a,b)\) < 1
= \(\frac{(a-b)^2}{(a+b)^2} + \frac{1}{(a+b)^2 }\)
= \(\frac{(a-b)^2+1}{(a+b)^2} \)
= \(\frac{(a^2+b^2-2ab +1)}{(a+b)^2} \)
= \(\frac{(a^2+b^2+2ab -4ab +1)}{(a+b)^2} \)
= \(\frac{(a+b)^2}{(a+b)^2} - \frac{(4ab-1)}{(a+b)^2}\)
= \(1 - \frac{(4ab-1)}{(a+b)^2}\)
= 1 - (positive real number).......... {as a and b are positive integers, 4ab-1 ≥ 4*1*1 -1=3}
\( F^2_{1} (a,b) + F_{2} (a,b)\) < 1
rheam25
21 Jan 2025, 07:29
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\( F^2_{1} (a,b) + F_{2} (a,b)\) = \( (\frac{(a-b)}{(a+b)})^2 + \frac{1}{(a+b)^2 }\)
= \( (\frac{(a-b)^2}{(a+b)^2} + \frac{1}{(a+b)^2 }\)
= \(\frac{(a-b)^2 + 1}{(a+b)^2}\)
Now, lets take some numbers to understand this
1. a = b= 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(1-1)^2 + 1}{(1+1)^2}\)
= \(\frac{(0)^2 + 1}{(2)^2}\) = \(\frac{1}{4}\) = 0.25
2. Lets take a bigger value of a and a smaller value of b. a = 9, b = 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(9-1)^2 + 1}{(9+1)^2}\)
= \(\frac{(8)^2 + 1}{(10)^2}\) = \(\frac{65}{100}\)
= 0.65
Now, value of the expression increased but it is still < 1
3. Lets take a huge value of a and small value of b. a = 10,000, b = 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(10000-1)^2 + 1}{(10000+1)^2}\)
= \(\frac{(9999)^2 + 1}{(10001)^2}\)
Now 1 + \((9999)^2 \) ~ \( (9999)^2 \) as 1 is very small number
~ \(\frac{(9999)^2}{(10001)^2}\)
Which is a bigger fraction but is still less than 1
So, numerator will always be lesser than the denominator and will be < 1
So, Answer will be A
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
= \( (\frac{(a-b)^2}{(a+b)^2} + \frac{1}{(a+b)^2 }\)
= \(\frac{(a-b)^2 + 1}{(a+b)^2}\)
Now, lets take some numbers to understand this
1. a = b= 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(1-1)^2 + 1}{(1+1)^2}\)
= \(\frac{(0)^2 + 1}{(2)^2}\) = \(\frac{1}{4}\) = 0.25
2. Lets take a bigger value of a and a smaller value of b. a = 9, b = 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(9-1)^2 + 1}{(9+1)^2}\)
= \(\frac{(8)^2 + 1}{(10)^2}\) = \(\frac{65}{100}\)
= 0.65
Now, value of the expression increased but it is still < 1
3. Lets take a huge value of a and small value of b. a = 10,000, b = 1
=> \(\frac{(a-b)^2 + 1}{(a+b)^2}\) = \(\frac{(10000-1)^2 + 1}{(10000+1)^2}\)
= \(\frac{(9999)^2 + 1}{(10001)^2}\)
Now 1 + \((9999)^2 \) ~ \( (9999)^2 \) as 1 is very small number
~ \(\frac{(9999)^2}{(10001)^2}\)
Which is a bigger fraction but is still less than 1
So, numerator will always be lesser than the denominator and will be < 1
So, Answer will be A
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
