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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