from the given,
if a= 1, then b = any integer from -9 to 9 except 0
if a = -1, then b = any even integer from -8 to 8 except 0

from statement (1),
if a =1, then b = 1 or -1
if a = -1, then b = any integer from -9 to 9 except 0 --> insufficient

from statement (2),
if a =1, then b = 1,5,6 (mono-cyclic integers) or -5 ( which leads to a mono-cyclic integer 5)
if a =-1, then b = 5 or -5 (which lead to a number with 5 as its unit digit) --> insufficient

upon combining,
if a = 1, then b = 1 --> sufficient C
if a = -1, then b has no valid common value (which means that a is not equal -1)

If a and b are non-zero single digit integers such that a^b=1, what is the value of b?

(1) −1≤b^a≤1
(2) 10a+b raised to any positive power always has its units digit as b.

Analysis of question stem tells us that a =1 since a is non zero integer and a^b =1.
As per option (1) b can be either -1 or 1 - hence not sufficient
As per option (2) values of b can be either 1,5 or 6 as 10a+b will always have unit digit of b and any power of b has to result in same unit digit as b. so only 3 values satisfy that. - not sufficient

Option (1) + (2) - sufficient as only number which satisfies both is 1. - hence Answer C

Bunuel: As per comments answer is C but right answer for this question is E. Can you please explain how?

@kitipriyanka,i am no expert but I could help

If a and b are non-zero single digit integers such that a^b=1 what is the value of b?

(1) −1≤b^a≤1
(2) 10a+b10a+b raised to any positive power always has its units digit as b.
Before we answer the question , what non zero integers will give us a^b =1.
Either 1 raised to the power any integer or -1 raised to an even integer


Statement 1 : a=1,b=1 works for us.a=-1 and b=6 works as well.Insufficient


statement 2 : Now we are told in the stem that, a and b are non zero digits. Meaning 10a+b will have its units digit as b.We know that numbers with the base of 0,1,5 or 6 when raised to a positive power will retain the units digit of the base.Obviously,b can not be 0 but it can be 1 or 5 or 6.In sufficient

together if b=6 and a =-1 ,we satisfy both statements and the question stem.If b=1 and a=1,we satisfy both statements and the question stem as well.In both cases b has 2 different values of 1 and 6.Insufficient

Answer is E

For statement (2), you almost have it right:
if a = 1 then b = 1,5,6 AND -4,-5,-9 since negative single digits are allowed
if a = -1 then b = 9,5,4 AND -1,-5,-6 since negative single digits are allowed

combining 1+2:
if a = 1, then b = 1
if a = -1, then b = -6 or 4
So now you have two choices for a itself, where b can be 1, -6 or 4
Insufficient

If a and b are non-zero single digit integers such that \(a^b\)=1, what is the value of b?

Second statement seems easier

(2) unit digit of \((10a+b)^x\) is b.
Possible values of b are 1, 5, 6. Insufficient

(1) −1≤\(b^a\)≤1
When b =1, a=-1
When b = 5, a = -1
value of b is still not known. Insufficient

(1)+(2), still b can be 1, 5 or 6
Insufficient

IMO E
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Question stem: \(a^b = 1\)

These are the possible values a and b can take to fulfil that condition:
a = -1, b = {any even value}
a = 1, b = {any vale}

Need to find the value of B!

1. b = 1 or -1, a = {any value}
a = -1, b= {any value}

not sufficient!

2. a = {any value}, b = 1 (because 1 raised to any power will ultimately have it unites digits as 1)
a = {any value}, b = 6 (because 6 raised to any power will ultimately have it unites digits as 6)

Combine 1&2
a=1, b=1
a= -1, b =6

NOT SUFFICIENT
E

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Because I found the answers above to be either incorrect or incomplete, I have tried to provide the most exhaustive solution I could come up with! Please let me know if there are any corrections to be made.

a and b are non-zero single-digit integers such that a^b=1

2 cases are possible:
(i) a = 1; => b can be any integer
(ii) a = -1; => b has to be an even integer

Statement 1:
-1 <= b^a <= 1

-> if a = 1
=> b = 1, -1

-> if a = -1
=> -1 <= 1/b <= 1
=> b can be -6,-4,-2,2,4,6,.....
eg: b = 2
=> -1 <= 0.5 <= 1

=> Many possible values of b

=> Statement 1 is insufficient

Statement 2:

-> if a = 1
=> b = 1,5,6

-> if a = -1
=> b = 4, -6

=> Many possible values of b

=> Statement 2 is insufficient


Statement 1 & Statement 2:

-> if a = 1
=> b = 1

-> if a = -1
=> b = 4,-6

=> Many possible values of b

=> Statement 1 and Statement 2 are insufficient

Answer: E
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In statement 1, when a = 1, how can b be 5?
Because, when a = -1, b = even integer, since a^b = 1.

Please correct me, if I'm wrong
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