Which of the following code snippets will print True to the monitor? (Choose two.)

The correct answer is A. ``` 1 print('is' in 'This IS Python code.') ``` D. ``` 1 print('t' in 'Peter') ```. Options A and D both use the in operator to test substring membership, which is case-sensitive: 'is' is found inside 'This...' and 't' is found inside 'Peter', so both return True. B is wrong because small integers like 42 are cached by CPython (integers -5 to 256 share the…

Question

Which of the following code snippets will print True to the monitor? (Choose two.)

Options

1 print('is' in 'This IS Python code.')

B
```
1 x = 42
2 y = 42
3 print(x is not y)
```

1 x = 'Peter Wellert'
2 y = 'Peter Wellert'.lower()
3 print(x is y)

D
```
1 print('t' in 'Peter')
```

1 x = ['Peter', 'Paul', 'Mary']
2 y = ['Peter', 'Paul', 'Mary']
3 print(x is y)

How the community answered

(28 responses)

A
71% (20)
B
14% (4)
C
4% (1)
E
11% (3)

Explanation

Options A and D both use the in operator to test substring membership, which is case-sensitive: 'is' is found inside 'Th**is**...' and 't' is found inside 'Pe**t**er', so both return True.

B is wrong because small integers like 42 are cached by CPython (integers -5 to 256 share the same object), so x is y is actually True, making x is not y → False. C is wrong because .lower() produces a new string object ('peter wellert'), which is a different object than x ('Peter Wellert'), so x is y is False. E is wrong because two separately created lists - even with identical contents - are always distinct objects in memory, so x is y is False.

Memory tip: Think of in as "is it inside?" (checks content/membership) and is as "is it the identical object?" (checks memory address). Confusing them is the core trap in this question - always reach for == to compare values and reserve is for identity checks like x is None.

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