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03 Jul 2012, 18:17
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:51) correct
65%
(02:36)
wrong
based on 418
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Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between
(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½
(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½
04 Jul 2012, 02:32
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Smita04
The question should read:
\(A_x(y)\) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then \(A_x(A_x(A_x(A_x(A_x(x)))))\) is between
(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½
According to the stem:
\(A_x(x)=(x+1)x=x^2+x\);
\(A_x(A_x(x))=A_x(x^2+x)=(x^2+x+1)*x=x^3+x^2+x\);
\(A_x(A_x(A_x(x)))=A_x(x^3+x^2+x)=(x^3+x^2+x+1)*x=x^4+x^3+x^2+x\);
...
We can see the pattern now, so \(A_x(A_x(A_x(A_x(A_x(x)))))=x^6+x^5+x^4+x^3+x^2+x\).
For \(x=\frac{2}{3}\) we'll get: \((\frac{2}{3})^6+(\frac{2}{3})^5+(\frac{2}{3})^4+(\frac{2}{3})^3+(\frac{2}{3})^2+(\frac{2}{3})\).
So, we have the sum of the 6 terms of the geometric progression with the first term equal to \(\frac{2}{3}\) and the common ratio also equal to \(\frac{2}{3}\).
Now, the sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term. So, if we had infinite geometric progression instead of just 6 terms then its sum would be \(Sum=\frac{\frac{2}{3}}{1-\frac{2}{3}}=2\). Which means that the sum of this sequence will never exceed 2, also as we have big enough number of terms (6) then the sum will be very close to 2, so we can safely choose answer choice D.
Answer: D.
One can also use direct formula.
We have geometric progression with \(b=\frac{2}{3}\), \(r=\frac{2}{3}\) and \(n=6\);
\(S_n=\frac{b(1-r^n)}{(1-r)}\) --> \(S_{6}=\frac{\frac{2}{3}(1-(\frac{2}{3})^{6})}{(1-\frac{2}{3})}=2*(1-(\frac{2}{3})^{6})\). Since \((\frac{2}{3})^{6}\) is very small number then \(1-(\frac{2}{3})^{6}\) will be less than 1 but very close to it, hence \(2*(1-(\frac{2}{3})^{6})\) will be less than 2 but very close to it.
Answer: D.
Hope it's clear.
General Discussion
03 Jul 2012, 21:51
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If I read the question correctly, AX(Y) = (1+Y)X
Putting X=2/3,
=> AX(X) = (1+ 2/3)*(2/3)
=> AX(X) = (5/3)*(2/3)
=> AX(X) = 10/9 > 1
AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2.
The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.
Regards,
Tushar
Putting X=2/3,
=> AX(X) = (1+ 2/3)*(2/3)
=> AX(X) = (5/3)*(2/3)
=> AX(X) = 10/9 > 1
AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2.
The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.
Regards,
Tushar
