If a + b > 0 and a^b < 0, which of the following must be true?

a^b<0 --> means a<0 (i)
a+b>0 and a<0 --> it must be true that b>0 and |b|>|a| (ii) and (iii).

(i), (ii) and (iii) must be true.

FINAL ANSWER IS (D)

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We are given that, a+b>0 and a^b<0,

From a^b<0, we can say that a is negative and b is odd number.
a<0, is ok
b>0, is ok
For being a+b>0, III |b|>|a| is true. Hence: (D)

In this question on inequalities, we can use some basic properties of inequalities and solve this question without too much of difficulty.

Let us try to analyse the data given in the question statement.

1) a+b>0. If the sum of two numbers is more than ZERO, there are not many conclusions we can draw from this situation about the signs of the numbers. The only things we can conclusively say are:

Both the numbers cannot be negative

Both the numbers cannot be ZERO

If a<0, b>0 in such a way that the absolute value of b is more than the absolute value of a.

If b<0, a>0 in such a way that the absolute value of a is more than the absolute value of b.

2) The second piece of information given in the question is \(a^b\)<0. This means that a is definitely negative and b is positive.

From the above, it’s clear that statement I and II are definitely true. Based on this, answer options A and C can be eliminated.

a is negative, b is positive and a+b>0 means that the absolute value of b is definitely more than the absolute value of a i.e. |b| > |a|. Statement III is also definitely true. Answer options B and E can be eliminated

The correct answer option is D.

As you see, there weren’t too many advanced concepts we applied here. We stuck to the basics and we were able to solve it in less than 2 minutes. This is probably why there is so much emphasis on being good with your basics, regardless of the topic.

Hope that helps!

a^b < 0 is only possible when
--> a < 0 & b is positive (either positive or negative)
E.g: \((-2)^3\), \((-2)^{-3}\), . . . .

If \(a + b > 0\) & if "\(a\)" is negative
--> "\(b\)" must be positive with greater numerical value than that of "\(a\)"
E.g: -2 + 3, -4 + 10 etc
--> |b| > |a| always

--> I & III must be true

Option C

a+b>0: a,b=++ or a,b=+-,-+
a^b<0: a,b≠0, a,b=-,+

I. true; II. true; III. true: a+b>0, if a,b=-,+, then |b|>|a| or else, a+b<0.

Ans (D)

The given condition (a^b > 0) is only possible when :-
a < 0 & b is positive or negative
E.G. \((-4)^5, (-4)^{-5},.....\)
.
.
.
If a + b > 0 and a is negative.
Then b is positive and b > a
.
Hence it won't be wrong to say.
\(|b| > |a|\)
.
And therefore I & III must be true.

.
Option C

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test the values :

a=-2 and b = 3
we get yes to all options
IMO D; all correct

If a+b>0and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|

A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these

If a+b>0 and \(a^b<0\), which of the following must be true?

I. a<0
II. b>0
III. |b|>|a|

A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these

\(a^b<0\) where a < 0 and b is odd always. Thus
I. a<0 always True
II. b>0 false when b is even
III. |b|>|a| False when b = - 2 and a = -2 even through \(a^b<0\)

Answer A.

If a+b > 0 and a^b < 0, which of the following must be true?

Constraint: a+b >0 and a^b < 0
Now if a^b <0 ,then a must be negative and b must be odd or cube root
Now if a+b>0, then b must be +ve since a is -ve and b is +ve odd or cube root
where a is not equal to b

I. a < 0 (Must be true)
II. b > 0 (Must be true)

III. |b| > |a|
When a = -2 ,b= 3 —-> a^b <0
and a=-2,b=3 —-> a+b>0
Yes |b| > |a| (Must be true)
Or
|b| > |a| says the distance of b from zero(0) is greater than the distance of a from zero(0)
<——a(-2)——0———b(3)——>
since a+b>0

A. I only
B. I and II only
C. I and III only
D. I, II ,and III
E. None of these

Hit that D

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If a+b >0 and \(a^{b}<0\), which of the following must be true?
--> In order \(a^{b}\) to be less then zero, --> a must be less than zero (a <0)

We got:
a+b >0 and -a >0
--> b>0

Also, we got:
b>0
-a >0
--> b-a >0

I. a<0
II. b>0
III. |b|>|a|
--> square the both sides--> \(b^{2}-a^{2} >0\)
(b-a)*(b+a) >0

All are TRUE.
The answer is D.

Ans: D
I, II and III
as a^b<0..so a must be negative ,so a<0
as a<0 and a+b>0, b must be positive b>0
III is also true

Archit3110

\(a^b\) for the chosen values (a=5 and b=-3) is NOT satisfied so the chosen values are incorrect to test the three conditions.

GMATinsight ; my bad i did an error

(1/a^x = a^-x) ( exponents)

silly error I did..

\(a + b > 0\) and \(a^b < 0\)

Possibilities for a + b > 0 are:

#1. \(a > 0\) and \(b > 0\)
=> fails the test for \(a^b < 0\) - Hence rejected

#2. \(a > 0\) and \(b < 0\), but \(|a| > |b|\) (i.e. \(a\) is a larger positive and \(b\) is a smaller negative)
=> fails the test for \(a^b < 0\) (since any positive number \(a\) raised to any exponent \(b\) remains positive) - Hence rejected

#3. \(a < 0\) and \(b > 0\), but \(|a| < |b|\) (i.e. \(b\) is a larger positive and \(a\) is a smaller negative)
=> passes the test for \(a^b < 0\) on condition that \(b\)\(\) is an odd number (since any negative number \(b\) raised to any odd exponent \(a\) remains negative) - Hence possible

Working with the statements:

I. \(a < 0\) - True
II. \(b > 0\) - True
III. \(|a| < |b|\) - True

Answer D

Bunuel Isn't the question more as "which of the following can be true"

Bunuel
A quick question.

a^b will be negative when a<0 (as stated in option I) and b>0(as stated in option II). But at the same time, b must also be odd power for a^b to remain negative. If b ends up being even, then the term a^b will become positive even if a is negative.

Based on this reasoning, I rejected option II.

Let me know if my understanding is wrong.

Does II say anything about even or odd nature of b? It simply says: b > 0.

From a^b < 0, we can only say that a < 0 and b is not even. But b can be odd, not an integer, positive, or negative.