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C
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:
45% (02:37) correct
55%
(02:45)
wrong
based on 411
sessions
History
Date
Time
Result
Not Attempted Yet
If \(\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0\) where a and b are integers,which of the following could be the value of b?
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
20 Dec 2012, 21:25
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mun23
For \(\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0\) to hold, \((ab)^2+3ab-18 = 0\)
You need 'a' to be an integer so put in the values of b to check whether you get integral values for 'a'
b = 1 => a^2 + 3a - 18 = 0 => (a + 6)(a - 3) = 0 => Integral values so acceptable
b = 2 => 4a^2 + 6a - 18 = 0 => (2a + 6)(2a - 3) = 0 => We get a = -3 (an integer) hence acceptable
b = 3 => 9a^2 + 9a - 18 = 0 => (a + 2)(a - 1) = 0 => We get a = -2 or 1. a can take neither of these values since they make the denominator 0. Not acceptable
Answer (C)
General Discussion
16 Dec 2012, 04:22
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Ans:
from the fraction we get that “a” cannot be 1 or -2, so putting the value of a in numerator we get the the values of “b” to 3,-6, -3/2 , therefore b cannot be 3 so the answer is (C).
from the fraction we get that “a” cannot be 1 or -2, so putting the value of a in numerator we get the the values of “b” to 3,-6, -3/2 , therefore b cannot be 3 so the answer is (C).
