It's an error in the MGMAT book. They have given it in their errata. The link of their errata is given in this post (on the previous page):
if-a-is-not-equal-to-zero-is-1-a-a-b-122266.html#p989860
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Thanks Karishma for clearing the doubts...Can you pls explain what does IaI = IbI mean....in simpler terms....

If b^4 + 3 > a^2
=> a^2 - b^4 < 3

Now

2. a^2 = b^4
=> a^2 -b^4 = 0
It tells a^2 - b^4 < 3 is sufficient

Wondering what wrong assumption I am making with above.

|a| = |b| means that the distance of a from 0 is equal to the distance of b from 0.
This means, if a = 5, b = 5 or -5
Similarly, if a = -5, b = 5 or -5

So, imagine the number line. There are two points at a distance of 5 from 0. a and b could lie on any one of these points.
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'Is \(\frac{1}{a} > \frac{a}{(b^4 + 3)}\)' is NOT the same as 'Is\(b^4 + 3 > a^2\)?'

Mind you, it is not given that 'a' is positive. You cannot cross multiply in an inequality if you do not know the sign of the varaible.

e.g.
a < b/c is not the same as ac < b

If we know that c is positive, then it is ok. Then a < b/c is same as ac < b
If instead, c is negative, then a < b/c is the same as ac > b (Note that the inequality sign has flipped)

Hence, statement (2) is not sufficient alone.
Statement (1) tells us the sign of a and we see that it is sufficient alone (check my solution above)
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1. If a NOT=0, is \(\frac{1}{a}>\frac{a}{b^4+3}\)

i. \(a^2=b^2\)
ii. \(a^2=b^4\)
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IMO E

LET a = 1
according to statement 1==> \(b^4=1\)
now putting in equation
\(\frac{1}{1}>\frac{1}{1+3}\)==>satisfies
if \(a = -1 b^4 =1\)
\(\frac{1}{-1}>\frac{-1}{1+3}\)==>doesnt satisfies.

statement 2:
\(a^2 = b^2\)
let \(a=1 b^4 = 1\)
\(\frac{1}{1}>\frac{1}{1+3}\)==>satisfies

let \(a=-1 b^4 = 1\)
\(\frac{1}{-1}>\frac{-1}{1+3}\)===>doesnt satisfies.

combining both
\(a,b= 1 or -1\)

same cases will not satisfy

hence E
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The question asks IS \(\frac{(b^4+3)}{a}>a \to\) Multiply both sides by\(a^2 \to a(b^4+3)>a^3 \to\) Is \(a(b^4+3-a^2)>0\).

From F.S 1, the question becomes: Is \(a(a^4-a^2+3)>0\).

Note that \((a^4-a^2+3)\) will always be positive as because it can be represented as sum of 2 squares : \((a^4-2a^2+1)+a^2+2 = (a^2-1)^2+(a^2+2)\)

Thus, the question now simply becomes : Is a>0

We clearly don't know that. Hence, Insufficient.

From F.S 2, the question becomes : Is \(a(b^4+3-a^2)>0 \to a(a^2+3-a^2)>0 \to\) Is \(3a>0\). Again, Insufficient.

Taking both together, we know that\(b^2 = b^4 \to b^2(b^2-1) = 0 \to b^2=a^2 = 1 [b^2 \neq{0}\) as\(a\neq{0}]\)
Thus, a could be\(\pm1\). Insufficient.

E.

Sidenote: The actual question is from Manhattan Gmat and they had published an errata for this question , which modifies the first fact statement to
\(a = b^2\). Now, if that would have been the case,the answer would have been A. However, with the given condition, the answer is E.
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Merging similar topics.
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Hi Karishma,
As per the above highlighted part, if \(a\) is '-ve' then \(a=-b^2\), so \(\sqrt{a}\) will be \(\sqrt{-b^2}\). Hence \(a\) becomes imaginary as \(b^2\) is always positive.So here also \(a\) must be '+ve'.

But, we've considered \(a\) as '-ve' also! Could you please explain this?

Much appreciate your feedback.
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\(a^2 = b^4\)

When you take the square root of both sides here, you get \(|a| = |b^2| = b^2\)
You do not get a = b^2.
Note that when you take square root of \(x^2 = y^2\), you get |x| = |y|, not x = y

So a can still be negative. Say a = -9, b = 3
In this case, \(a^2 = b^4 = (-9)^2 = 3^4\)
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Thanks Karishma for the explanation and apologies for late acknowledgement as I could figure it out later and hence this post slipped somehow

But, again much thanks. +1
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