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| Certification Vendor: | Salesforce |
|---|---|
| Exam Name: | Salesforce Certified JavaScript Developer - Multiple Choice |
| Exam Number: | JS-Dev-101 |
| Related Certifications: | Salesforce Certified Platform Developer I Salesforce Certified Developer |
| Exam Price: | USD 200 |
| Exam Format: | Multiple choice, Multiple select |
| Certificate Validity Period: | Does not expire; requires maintenance updates |
| Exam Duration: | 105 minutes |
| Available Languages: | English, Japanese |
| Real Exam Qty: | 60 scored + up to 5 unscored |
| Passing Score: | 65% |
| Recommended Training: | Trailhead JavaScript Developer I Certification Prep |
| Exam Registration: | Salesforce Certification Registration |
| Sample Questions: | Salesforce JS-Dev-101 Sample Questions |
| Exam Way: | Online proctored or onsite at authorized testing centers |
| Pre Condition: | No formal prerequisites; recommended: strong JavaScript fundamentals, ES6+ experience, web development familiarity |
| Official Syllabus URL: | https://trailheadacademy.salesforce.com/certificate/exam-javascript-dev---JS-Dev-101 |
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NEW QUESTION # 99
Which two code snippets show working examples of a recursive function?
Answer: A,C
Explanation:
return startNumber;
}
};
(Note: Option D is shown here with corrected syntax: lowercase return and matching parentheses.) Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
A recursive function is a function that calls itself and has a base case to terminate the recursion.
Evaluate each option:
Option A:
const sumToTen = numVar => {
if (numVar < 0)
return;
return sumToTen(numVar + 1);
};
This function calls itself: sumToTen(numVar + 1) - so it is recursive.
However, the base condition is if (numVar < 0) return;.
If you call sumToTen(0):
numVar < 0 is false, so it calls sumToTen(1), then sumToTen(2), and so on, incrementing forever.
There is no condition to stop the recursion when numVar increases; it will eventually cause a stack overflow.
This code does not represent a properly working recursive function with a valid termination for increasing values and is not a good example of correct recursion.
Option B:
function factorial(numVar) {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar - 1;
}
This function does not call itself anywhere.
It has conditional returns, but there is no recursive call such as factorial(numVar - 1).
Therefore, it is not recursive at all.
Option C:
const factorial = numVar => {
if (numVar < 0) return;
if (numVar === 0) return 1;
return numVar * factorial(numVar - 1);
};
This is a classic recursive factorial implementation.
It calls itself with a smaller argument: factorial(numVar - 1).
Base cases:
If numVar < 0, it simply returns (could be treated as invalid input).
If numVar === 0, it returns 1, which is the mathematical definition of 0! (zero factorial).
For positive integers, it correctly multiplies numVar by factorial(numVar - 1) until it reaches the base case.
This is a correct and working recursive function.
Option D (corrected):
let countingDown = function(startNumber) {
if (startNumber > 0) {
console.log(startNumber);
return countingDown(startNumber - 1);
} else {
return startNumber;
}
};
This function also calls itself: countingDown(startNumber - 1).
Base case:
When startNumber is not greater than 0 (i.e., 0 or negative), it returns startNumber and stops recursing.
For example, countingDown(3) would:
Log 3, call countingDown(2)
Log 2, call countingDown(1)
Log 1, call countingDown(0)
At 0, it hits the else branch and returns 0, ending the recursion.
This is a valid working recursive function structure (once syntax is corrected).
Therefore, the snippets that show working recursive functions are:
Answe r: C, D
Study Guide / Concept Reference (no links):
Definition of recursion: a function calling itself
Base case vs recursive step
Recursive factorial implementation
Recursive countdown example
Importance of a terminating condition to avoid infinite recursion
NEW QUESTION # 100
JavaScript:
01 function Tiger() {
02 this.type = 'Cat';
03 this.size = 'large';
04 }
05
06 let tony = new Tiger();
07 tony.roar = () => {
08 console.log('They\'re great!');
09 };
10
11 function Lion() {
12 this.type = 'Cat';
13 this.size = 'large';
14 }
15
16 let leo = new Lion();
17 // Insert code here
18 leo.roar();
Which two statements could be inserted at line 17 to enable line 18?
Answer: C,D
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge There are two valid ways to ensure leo.roar() exists:
Directly assign a roar function to leo (Option A).
Assigning a property to an instance object creates a new method on that instance.
leo.roar = () => { console.log('They\'re pretty good!'); };
After this, calling leo.roar() is valid.
Copy tony's properties into leo using Object.assign (Option B).
Object.assign(target, source) copies enumerable own properties from the source object into the target object.
Since tony.roar exists, executing:
Object.assign(leo, tony);
copies roar into leo, making leo.roar() valid.
Why the other answers are incorrect:
Option C:
Object.assign(leo, Tiger) copies properties from the function object Tiger, not from Tiger.prototype and not from a Tiger instance. Tiger (the function) has no roar property, so nothing useful is copied.
Option D:
leo.prototype is undefined because leo is an instance, not a constructor function. Only constructor functions have a .prototype property. This line would cause an error.
________________________________________
JavaScript Knowledge Reference (text-only)
Instances have their own properties and do not contain a .prototype property.
Object.assign(target, source) copies own enumerable properties of the source object.
Assigning a function as a property of an object creates a callable method.
NEW QUESTION # 101
Why does second have access to variable a?
Answer: A
Explanation:
The variable a is declared in the outer function's scope. An inner function (second) forms a closure over its outer lexical scope, so it can access a. The reason is the scope chain / outer function's scope, not prototypes or hoisting.
NEW QUESTION # 102
Refer to thefollowing code that imports a module named utils:
import (foo, bar) from '/path/Utils.js';
foo() ;
bar() ;
Which two implementations of Utils.js export foo and bar such that the code above runs without error?
Choose 2 answers
Answer: A,D
NEW QUESTION # 103
Refer to the code below:
01 funetion Person() (
02 thie. fixsttane = "Zoha'y
os}
05 Perscn.prototype = [
06 jek: x => 'Developer'
onde
09 const myFather = new Peracn():
10 const result = myFather.firstiame + ' * + myFather.jeb()
What is the value of result after line 10 executes?
Answer: D
NEW QUESTION # 104
......
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