Salesforce JS-Dev-101 Exam – Complete Guide, Topics & Preparation Strategy
Are you planning to earn the Salesforce JS-Dev-101 certification? This comprehensive guide covers the Salesforce JS-Dev-101 exam overview, topics covered, key skills measured, fast facts, and preparation strategy to help you pass on your first attempt.
The Salesforce JavaScript Developer I (JS-Dev-101) exam validates your knowledge of JavaScript fundamentals, ES6+, Lightning Web Components (LWC), Salesforce platform integration, and modern web development best practices. This certification is ideal for developers building applications on the Salesforce platform.
Salesforce JS-Dev-101 Exam Overview
The Salesforce JS-Dev-101 exam is designed for candidates who want to demonstrate their expertise in JavaScript development within the Salesforce ecosystem.
Who Should Take the Salesforce JS-Dev-101 Exam?
This certification is ideal for:
Salesforce Developers
JavaScript Developers transitioning to Salesforce
Lightning Web Components (LWC) Developers
Front-end Developers working with Salesforce
Salesforce Platform App Builders
Technical Consultants implementing Salesforce solutions
If you are working with Lightning Web Components, Apex integration, REST APIs, and modern JavaScript frameworks, this exam is highly recommended.
Salesforce JS-Dev-101 Fast Facts
Exam Name: Salesforce JavaScript Developer I
Exam Code: JS-Dev-101
Number of Questions: 60 multiple-choice and multi-select questions
Pilot Questions: Up to 5 unscored questions
Passing Score: 65%
Exam Duration: 105 minutes
Delivery Method: Proctored exam (online or testing center)
Understanding these fast facts helps candidates plan their time management strategy effectively during the real test.
Topics Covered in Salesforce JS-Dev-101 Exam
The Salesforce JS-Dev-101 certification exam measures your knowledge across the following core domains:
1. JavaScript Fundamentals
Variables, data types, and operators
Functions and scope
Arrays and objects
Control flow statements
Error handling
2. Advanced JavaScript (ES6+)
Arrow functions
Promises and async/await
Destructuring
Spread and rest operators
Modules and imports
3. Lightning Web Components (LWC)
Component lifecycle
Data binding
Event handling
@wire service
Apex integration
Component communication
4. Salesforce Platform Integration
Working with Apex classes
Calling REST APIs
Handling server responses
CRUD operations
Security best practices
5. Browser and Web APIs
DOM manipulation
Fetch API
JSON handling
Client-side storage
6. Testing & Debugging
Debugging techniques
Browser developer tools
Best practices for performance optimization
Mastering these domains is essential to pass the Salesforce JavaScript Developer I certification exam successfully.
Why Get Salesforce JS-Dev-101 Certified?
Earning the Salesforce JS-Dev-101 certification demonstrates:
Strong knowledge of JavaScript for Salesforce development
Expertise in Lightning Web Components
Ability to build scalable Salesforce applications
Improved career opportunities in Salesforce ecosystem
Higher earning potential as a certified Salesforce developer
With Salesforce being one of the world’s leading CRM platforms, certified developers are in high demand globally.
How to Prepare for Salesforce JS-Dev-101 Exam
Preparation requires:
Strong understanding of JavaScript ES6+
Hands-on practice with Lightning Web Components
Experience integrating Apex and APIs
Mock exams and real exam simulation practice
Using reliable study material and exam practice tools significantly increases your chances of passing on the first attempt.
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Our Salesforce JS-Dev-101 practice tests and training PDF are designed to simulate the real exam environment, helping you build confidence and improve your time management skills.
Final Thoughts
The Salesforce JS-Dev-101 certification is a valuable credential for developers working with JavaScript and Lightning Web Components on the Salesforce platform. By mastering exam topics and practicing with reliable preparation material, you can confidently achieve success.
Start your Salesforce JS-Dev-101 exam preparation today and take the next step toward becoming a certified Salesforce JavaScript Developer!
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Sample Question and Answers
QUESTION 1
A team at Universal Containers works on a big project and uses yarn to manage the project’s dependencies.
A developer added a dependency to manipulate dates and pushed the updates to the remote repository.
The rest of the team complains that the dependency does not get downloaded when they execute yarn.
What could be the reason for this?
A. The developer added the dependency as a dev dependency, and YARN_ENV is set to production.
B. The developer missed the option –save when adding the dependency.
C. The developer added the dependency as a dev dependency, and NODE_ENV is set to production.
D. The developer missed the option –add when adding the dependency.
Answer: C
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge
In JavaScript server-side development using Node.js, dependency management is typically handled
through package managers such as npm or yarn. These tools categorize installed packages into:
dependencies “ required for running the application in any environment
devDependencies “ required only during development (testing tools, build tools, documentation
generators, etc.)
When a package is installed using:
yarn add <package> –dev
it is placed under the “devDependencies” section of package.json.
________________________________________
Behavior of Production Mode
Node.js uses the environment variable:
NODE_ENV=production
When this environment variable is set to production, both npm and Yarn follow the standard Node.js
convention and skip installing devDependencies. This is done to optimize production builds and
reduce deployment size. This is a known and documented behavior in Node.js package management tools.
Therefore, if:
The developer added the date-manipulation library as a dev dependency, and
Other team members execute yarn in an environment where NODE_ENV=production is set,
then Yarn will not install that dependency because devDependencies are intentionally excluded in production mode.
This explains the behavior described in the question.
________________________________________
Why the Other Options Are Incorrect
Option A:
“YARN_ENV is set to production” is incorrect because Yarn does not use the variable YARN_ENV for
dependency installation behavior. Node.js tools use NODE_ENV, not YARN_ENV.
Option B:
This is incorrect because Yarn automatically writes dependencies into package.json. Unlike older
npm versions, there is no need for the –save flag.
Option D:
There is no such option as –add. The correct syntax is simply:
yarn add <package>
Missing an option that does not exist cannot be the cause.
________________________________________
JavaScript Knowledge Reference (Text-Based, No Links)
Node.js uses the environment variable NODE_ENV to determine production or development mode.
Package managers (npm and Yarn) follow the rule that when NODE_ENV=production, only
“dependencies” are installed and “devDependencies” are skipped.
Yarn automatically persists installed packages to package.json without requiring –save.
Yarn uses the command yarn add to add dependencies; there is no –add flag.
QUESTION 2
Refer to the code below:
01 let o = {
02 get js() {
03 let city1 = String(‘St. Louis’);
04 let city2 = String(‘New York’);
05
06 return {
07 firstCity: city1.toLowerCase(),
08 secondCity: city2.toLowerCase(),
09 }
10 }
11 }
What value can a developer expect when referencing o.js.secondCity?
A. undefined
B. An error
C. ‘New York’
D. ‘new york’
Answer: D
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge
1. Getter Functions in JavaScript
In JavaScript, when an object uses the get keyword, it defines a getter method. Accessing a getter
property executes the function and returns its value. Thus:
o.js
does not return the getter function; instead, it executes the function located at:
get js() { … }
and returns the object inside the return block.
________________________________________
2. Behavior of String() and toLowerCase()
Inside the getter:
let city1 = String(‘St. Louis’);
let city2 = String(‘New York’);
String() creates a string value.
Then, the returned object is constructed as:
{
firstCity: city1.toLowerCase(),
secondCity: city2.toLowerCase(),
}
The method toLowerCase() is a standard JavaScript string method that returns a new string with all
alphabetic characters converted to lowercase.
Therefore:
city2.toLowerCase()
returns:
‘new york’
________________________________________
3. Referencing the Property
When the developer writes:
o.js.secondCity
the following happens:
The getter js runs and returns an object.
The returned object includes the property:
secondCity: ‘new york’
Accessing .secondCity retrieves the lowercase string ‘new york’.
Therefore, the correct value is ‘new york’.
________________________________________
Why the Other Options Are Incorrect
A . undefined “ incorrect because the property secondCity clearly exists in the returned object.
B . An error “ incorrect because no invalid operations occur; all methods and properties are valid.
C . ‘New York’ “ incorrect because toLowerCase() transforms the string to lowercase.
________________________________________
JavaScript Knowledge Reference (Text-Based)
Getter methods using the get keyword return computed values when accessed.
JavaScript String values support the toLowerCase() method, which returns a lowercase version of the
original string.
Accessing nested properties like o.js.secondCity triggers the getter, returning the constructed object.
QUESTION 3
Refer to the code below:
01 x = 3.14;
02
03 function myFunction() {
04 ‘use strict’;
05 y = x;
06 }
07
08 z = x;
09 myFunction();
Considering the implications of ‘use strict’ on line 04, which three statements describe the execution of the code?
A. ‘use strict’ is hoisted, so it has an effect on all lines.
B. z is equal to 3.14.
C. ‘use strict’ has an effect between line 04 and the end of the file.
D. ‘use strict’ has an effect only on line 05.
E. Line 05 throws an error.
Answer: B, D, E
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
Behavior of non-strict global code
The script does not begin with a ‘use strict’ directive at the top level, so the global code (outside any
function) runs in non-strict (sloppy) mode.
Line 01: x = 3.14;
In non-strict mode, assigning to an undeclared identifier (no var, let, or const) creates an implicit
global variable. So after line 01, x exists and equals 3.14.
Line 08: z = x;
This also runs in non-strict mode. Since x is already defined (from line 01), z is set to 3.14. Therefore,
statement B is correct: z is equal to 3.14.
Scope of ‘use strict’ inside a function
The line:
‘use strict’;
inside myFunction is a directive prologue for that function, not for the entire file. This means:
Strict mode applies only within the body of myFunction, from the directive to the end of that
function.
It does not retroactively affect code before the function or code outside of it.
Therefore:
Statement A is incorrect: ‘use strict’ is not oehoisted to affect the whole file. It only affects the
function body.
Statement C is incorrect: strict mode does not apply from line 04 to the end of the file; it only applies
within myFunction, not to global lines like 01, 08, or 09.
Execution of myFunction in strict mode
When myFunction is called on line 09:
function myFunction() {
‘use strict’;
y = x;
}
Within this function:
Strict mode is active for its body.
In strict mode, assigning to an undeclared variable (like y here) is not allowed and results in a
ReferenceError at runtime.
So line 05:
y = x;
throws a ReferenceError because y is not declared with var, let, or const.
Therefore, statement E is correct: line 05 throws an error.
Why statement D is considered correct
Statement D says:
‘use strict’ has an effect only on line 05.
In terms of execution in this specific code:
The only executable statement in myFunction that is affected by strict mode is the assignment on line
05.
The directive itself on line 04 is not a oenormal runtime operation; it is a directive that sets the mode.
There are no other statements inside the function that behave differently under strict mode; only the
line that assigns to the undeclared variable shows a strict-mode effect.
So, describing the practical execution behavior of this snippet, strict mode manifests its effect only on
line 05, making statement D correct in this context.
Summary of each option:
A: Incorrect “ strict does not apply to all lines in the file.
B: Correct “ z becomes 3.14 in non-strict global code.
C: Incorrect “ strict does not affect code outside the function.
D: Correct “ in this code, the only behavioral effect of strict mode is on line 05.
E: Correct “ line 05 throws a ReferenceError due to assignment to undeclared y under strict mode.
JavaScript knowledge references (descriptive, no links):
In non-strict (sloppy) mode, assigning to an undeclared identifier creates a global variable.
‘use strict’ inside a function body enables strict mode for that function only.
In strict mode, assigning to an undeclared variable results in a ReferenceError.
Directive prologues (‘use strict’) affect only their containing function or script, not other scopes.
==================================================
QUESTION 4
Refer to the code below:
01 let total = 10;
02 const interval = setInterval(() => {
03 total++;
04 clearInterval(interval);
05 total++;
06 }, 0);
07 total++;
08 console.log(total);
Considering that JavaScript is single-threaded, what is the output of line 08 after the code executes?
A. 11
B. 12
C. 10
D. 13
Answer: A
Explanation:
Comprehensive and Detailed Explanation From Exact Extract JavaScript knowledge:
Synchronous execution order
JavaScript executes code in a single thread, following a well-defined order:
All synchronous code runs first, line by line.
Asynchronous callbacks (like those scheduled with setInterval or setTimeout) are placed into the
event queue and executed only after the current call stack is empty.
Lets follow the code step by step:
Line 01:
let total = 10;
total is initialized with the value 10.
Line 02:
const interval = setInterval(() => {
total++;
clearInterval(interval);
total++;
}, 0);
setInterval schedules the callback function to run repeatedly after a delay of at least 0 milliseconds,
but it does not run immediately. The callback is added to the timer queue and will be invoked after
the current synchronous script finishes and the event loop gets to process timer callbacks.
At this point, interval holds the interval ID, but the callback has not executed yet.
Line 07:
total++;
This is still synchronous, so it runs before any scheduled callbacks.
total was 10, now it becomes 11.
Line 08:
console.log(total);
At this moment, the interval callback has still not run (because the event loop has not yet processed
the timer queue).
So total is 11, and console.log(total); outputs 11.
Therefore, the value printed at line 08 is 11, making option A correct.
What happens after the log (for understanding, not affecting the answer)
After the main script finishes, the event loop processes the timer callback for setInterval:
Callback:
() => {
total++; // from 11 to 12
clearInterval(interval); // cancels further executions
total++; // from 12 to 13
}
So eventually total becomes 13, but this happens after console.log(total) has already executed. Since
the question asks specifically for the output at line 08, the asynchronous updates do not change that
lines output.
Why other options are incorrect
Option B (12): This would require the callback to run before the log, which does not happen because
asynchronous callbacks are queued and executed after the current stack finishes.
Option C (10): Ignores the total++ on line 07.
Option D (13): This is the final value after the callback finishes, but it occurs after the console.log line
executes, not at the time line 08 runs.
JavaScript knowledge references (descriptive, no links):
JavaScript is single-threaded and uses an event loop with a call stack and task queues.
setInterval schedules callbacks to run asynchronously after a minimum delay; the callback never runs
before the current synchronous code finishes.
Synchronous statements like total++ on line 07 execute before any queued interval callback.
QUESTION 5
Refer to the code below:
01 const objBook = {
02 title: ‘JavaScript’,
03 };
04 Object.preventExtensions(objBook);
05 const newObjBook = objBook;
06 newObjBook.author = ‘Robert’;
What are the values of objBook and newObjBook respectively?
A. { title: “JavaScript” }
{ title: “JavaScript” }
B. { author: “Robert”, title: “JavaScript” }
undefined
C. { author: “Robert” }
{ author: “Robert”, title: “JavaScript” }
D. { author: “Robert”, title: “JavaScript” }
{ author: “Robert”, title: “JavaScript” }
Answer: A
Explanation:
________________________________________
Comprehensive and Detailed Explanation From Exact Extract JavaScript Knowledge
Object.preventExtensions(obj)
This built-in JavaScript method marks an object so that no new properties can be added to it.
Existing properties can still be read and updated, but adding new ones is disallowed.
const newObjBook = objBook;
Both variables reference the same object in memory. JavaScript objects are assigned by reference,
not copied.
newObjBook.author = “Robert”;
Because the object has been marked as non-extensible, JavaScript will not allow new properties to
be added.
The behavior depends on mode:
In non-strict mode: the assignment silently fails and does nothing.
In strict mode: this would throw a TypeError.
Since nothing indicates strict mode, this is non-strict behavior, making the assignment fail silently.
Therefore, the object remains:
{ title: “JavaScript” }
Both objBook and newObjBook point to the same unchanged object.
This matches option A.
________________________________________
JavaScript knowledge references (text-only)
Object.preventExtensions() prevents adding new properties.
Assigning an object to another variable copies the reference, not the object.
Adding a property to a non-extensible object silently fails in non-strict mode.
==================================================