The main confusion here is because the brackets are not specified in Quantity B. I fell into this trap too.
Quantity B is supposed to be: -(2^2x) and not (-2)^2x. This makes a huge difference, because irrespective of x, this quantity is going to remain negative.

The main counter argument to those arguing that the absence of brackets introduces a lot of ambiguity is that Quantity A DOES USE BRACKETS. (-3)^x is shown explicitly. Which means if they intended Quantity B to be (-2)^2x, they would have specified the brackets.

I think the answer is B as x is odd integar..and also i couldnot understand the power for -2 is (2X) or only X

x is a positive, odd integer.

Quantity A: (-3)^x
Quantity B: -2^(2x)

Note the difference in the 2 expressions:

Qty A: the power of x is applicable on (-3)
Qty B: the power of 2x is applicable on 2 and then the result is negated

Since x is odd: (-3) is raised to an odd power
=> [(-1)*(3)] is raised to an odd power

Thus, \((-1)^x = -1\) and 3^x remains as it is

Hence: Quantity A: \(-3^x\)

Also, Quantity B: -2^(2x) = -1 * 2^(2x) = -1 * (2^2)^x = -1 * 4^x = \(- 4^x\)

Thus, we now need to compare: \(- 3^x\) and \(- 4^x\)


Clearly, \(3^x < 4^x\) since x is positive integer

=> \(- 3^x > - 4^x\) (inequality reverses on multiplying a negative)


Answer A

ans is a because in qt.b is -1*some value by substituting value of x we will get some value which is multiplied by -1 later on

This is a common mistake which even I was about to make. So nope because -2^(2x) becomes -4^x and since x will always be a positive odd integer option B will always be greater.

I plugged 3, and then 1. Both times A > B, and because the plug-in numbers are narrowed down to positive odd numbers, there is no change in this trend.

(-3)^3 = -27 > -(2^2(3)) = -64

(-3)^1 = -3 > (-2)^2 = -4

In the case of Quantity A, \((-3)\) is raised to the power of \(x\), which is positive and odd. Since the odd powers preserve the sign of the base, Quantity A will always be negative

Let us take both conflicting cases

In the case of Quantity B, \(2\) is raised to the power \(2x\), and then the result is negated by the negative sign outside the bracket \(-(2^{2x})\). In this case the smaller of two adjacent numbers when raised to twice the power of the larger one, will be much bigger than the other one. That is \(2^{2x}\) will be many times greater than \(3^x\). If this isn't obvious, you can test it by letting \(x=1,3\) and \(5 \) and check. Thus, Quantity A is greater since we are computing \(-(2^{2x})\) and \((-3)^x\).

OR

In the case of Quantity B, \(2\) is raised to the power \(2\), and the result \(4\) is raised to the power \(x\). In this case it is obvious that between \((-3)^x\) and \(-(4^x)\), the latter will be smaller than the former, so Quantity A is the greatest.

So whatever is your interpretation of \(2^{2x}\), the answer is Quantity A.

No. Its -(2^2x) and not (-2)^2x.