mun23 wrote:
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
For the \(\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0\) = 0
numerator has to be zero
and a # 1 and a # -2 (Since these two values would make denominator zero and fraction undefined)
Now \((ab)^2+3ab-18\)
Put ab = x
\(x^2 + 3x - 18\)
factorized to
(x-3)(x+6)
Case I
x = ab = 3, => a =1 , b = 3 -- NOT POSSIBLE as we cannot have a = 1
x = ab = 3, => a =3 , b = 1 --
POSSIBLECase II
x = ab = -6, => a = -1, 2,
+3,
+6 & b =
+2, -3,
+6
Hence we can note that b can assume values 1 & 2 but not 3.
Hence C
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