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Salesforce JS-Dev-101 Exam Questions in exam preparation

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Salesforce Certified JavaScript Developer - Multiple Choice Sample Questions (Q120-Q125):

NEW QUESTION # 120
Given the code:
const copy = JSON.stringify([new String('false'), new Boolean(false), undefined]); What is the value of copy?

A. '[false, {}]'
B. '["false", false, null]'
C. '["false", false, undefined]'
D. '["false", {}]'

Answer: B

Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge JSON.stringify() applies the following rules:
String objects (new String('false'))
Are converted to their primitive string values.
So:
new String("false") → "false"
Boolean objects (new Boolean(false))
Convert to their primitive boolean value:
new Boolean(false) → false
Undefined values inside arrays
According to the JSON specification, undefined becomes null when stringifying an array:
JSON.stringify([undefined]) → "[null]"
Therefore, the array becomes:
["false", false, null]
And JSON.stringify wraps it in quotes:
'["false", false, null]'
This is exactly option A.
________________________________________
JavaScript Knowledge Reference (text-only)
Object wrappers (String, Boolean) are converted to primitives during JSON serialization.
undefined becomes null inside arrays when stringified.
JSON.stringify() follows the JSON data-model rules strictly.

NEW QUESTION # 121
Refer to the string below.
Const str='Salesforce';
Which two statementsresults in the word 'Sales'?

Answer:

Explanation:
See the Answer below in explanation

NEW QUESTION # 122
Which statement accurately describes an aspect of promises?

A. .then() cannot be added after a catch.
B. Arguments for the callback function passed to .then() are optional.
C. In a.then() function, returning results is not necessary since callbacks will catch the result of a previous promise.
D. .then() manipulates and returns the original promise.

Answer: B

NEW QUESTION # 123
A developer wants to create an object from a function in the browser using the code below.

What happens due to the lack of the mm keyword on line 02?

A. window.m Is assigned the correct object.
B. The m variable is assigned the correct object but this.name remains undefined.
C. window.name is assigned to 'hello' and the variable = remains undefined.
D. The m variable is assigned the correct object.

Answer: C

NEW QUESTION # 124
Refer to the code below:
01 const addBy = ?
02 const addByEight = addBy(8);
03 const sum = addByEight(50);
Which two functions can replace line 01 and return 58 to sum?

A. (Corrected for typing errors)
const addBy = (num1) => {
return function(num2) {
return num1 + num2;
}
B. const addBy = function(num1) {
return num1 * num2;
}
C. const addBy = (num1) => num1 + num2;
D. const addBy = function(num1) {
return function(num2) {
return num1 + num2;
}
}

Answer: A,D

Explanation:
}
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
We want:
const addByEight = addBy(8);
const sum = addByEight(50);
And we need sum to be 58.
That means:
addBy(8) must return a function.
That returned function, when called with 50, must compute 8 + 50.
So addBy must be a higher-order function that returns another function capturing num1 and later using it with num2 (a closure).
Option A
const addBy = function(num1) {
return function(num2) {
return num1 + num2;
}
}
Step-by-step:
Call addBy(8):
num1 is 8.
The function returns an inner function: function(num2) { return num1 + num2; }.
So addByEight becomes this inner function, with num1 closed over as 8.
Then call addByEight(50):
num2 is 50.
The body computes num1 + num2, i.e., 8 + 50 = 58.
Therefore, with Option A in place:
const addByEight = addBy(8); // returns inner function
const sum = addByEight(50); // 58
So sum is 58. Option A is correct.
Option D (corrected)
const addBy = (num1) => {
return function(num2) {
return num1 + num2;
}
}
This is essentially the same logic expressed with an arrow function for the outer function:
Call addBy(8):
num1 is 8.
Returns the inner function function(num2) { return num1 + num2; }.
Call addByEight(50):
num2 is 50.
Computes num1 + num2 → 8 + 50 = 58.
So again:
const addByEight = addBy(8); // inner function with num1 = 8
const sum = addByEight(50); // 58
Option D is also correct.
Thus, A and D are the two functions that satisfy the requirement.
Why B and C are incorrect
Option B:
const addBy = function(num1) {
return num1 * num2;
}
This does not return a function; it returns a value.
addBy(8) returns 8 * num2, but num2 is not defined in this scope, which would cause a ReferenceError.
Also, addByEight would be a number (if num2 existed), not a function, so addByEight(50) would fail.
Option C:
const addBy = (num1) => num1 + num2;
Again, addBy returns a value, not a function.
addBy(8) returns 8 + num2, but num2 is not defined, so this is also invalid due to ReferenceError.
addByEight would be a number (or error), not a function.
Neither B nor C creates the required closure nor returns a function to be called later.
Reference / Study Guide concepts (no links):
Higher-order functions in JavaScript
Closures: inner functions capturing outer variables (num1)
Arrow functions vs function expressions
Returning functions from functions (function factories / currying)
Scope and ReferenceError when a variable is not defined
________________________________________

NEW QUESTION # 125
......

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