Problem 1 :
Let α and β be the roots of the quadratic equation
x^{2} + 2bx + b = 1
The smallest possible value of (α - β)^{2}
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4
Solution :
x^{2} + 2bx + b - 1 = 0
(α - β)^{2 }= (α + β)^{2} - 4 α β ---(1)
a = 1, b = 2b and c = b - 1
α + β = -b/a
α β = c/a
α + β = -2b/1 ==> -2b
α β = (b - 1)/1 ==> b - 1
(α - β)^{2 }= (-2b)^{2} - 4(b - 1)
= 4b^{2} - 4b + 4
= 4(b^{2} - b + 1)
It will have minimum (Smallest value) when x = 1/2.
4[(b - (1/2))^{2} + 3/4] ≥ 3
So, option D is correct.
Problem 2 :
It is known that n^{2012} + n^{2010} is divisible by 10 for some positive integer n. Which of the following numbers is not a possible value for n?
(A) 2 (B) 13 (C) 35 (D) 47 (E) 59
Solution :
= n^{2012} + n^{2010}
= n^{2010}(n^{2} + 1)
If the number is divisible by 10, then it should end with 0.
Option A : If n = 2
Cyclicity of 2 is 4.
2010/4 = remainder 2
2^{2} = 4.Unit digit of 2^{2010} is 4
= 2^{2010}(2^{2} + 1)
= 4(2^{2} + 1)
= 4(5)
Will end with 0, divisible by 10.
Option B : If n = 13
Cyclicity of 3 is 4.
2010/4 = remainder 2
3^{2} = 9.Unit digit of 3^{2010} is 9
= 9(13^{2} + 1)
= 9(unit digit of 13^{2} is 9 + 1)
= 9 (unit digit is 0)
Will end with 0, divisible by 10.
Option C : If n = 35
Cyclicity of 5 is 1.
2010/1 = remainder 0
5^{0} = 5.Unit digit of 5^{2010} is 5.
= 5(35^{2} + 1)
= 5(unit digit of 35^{2 }is 5 + 1)
= 5 (unit digit is 6)
Will end with 0, divisible by 10.
Option D : If n = 47
Cyclicity of 7 is 4.
2010/4 = remainder 2
7^{2} = 49.Unit digit of 7^{2010} is 9.
= 9(47^{2} + 1)
= 9(unit digit of 47^{2} is 9 + 1)
= 9 (unit digit is 0)
Will end with 0, divisible by 10.
Option E : If n = 59
Cyclicity of 9 is 2.
2010/2 = remainder 0
9^{0} = 1.Unit digit of 59^{2010} is 1.
= 1(59^{2} + 1)
= 1(unit digit of 59^{2} is 1 + 1)
= 9 (unit digit is 2)
Will not end with 0, not divisible by 10. So, option E is correct.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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