Pass the Salesforce Salesforce Developer JavaScript-Developer-I Questions and answers with CertsForce

Behavior of Promise.prototype.finally:

.finally(handler) registers a callback that runs when the promise is settled , meaning:

Either fulfilled (resolved) , or

rejected .

Important points:

The finally callback does not receive the promise’s value or error (unlike then and catch).

It is executed after the promise is settled, but before the resolution value or rejection reason is passed further down the chain.

It runs in both success and failure paths.

In the given code:

The promise may either:

Call reject( ' fraction > 0.5, ' + fraction) if fraction > 0.5, or

Call resolve(fraction) otherwise.

In both cases:

If it resolves , .then(() = > console.log( ' resolved ' )) runs, and then .finally(...) is executed.

If it rejects , .catch((error) = > console.error(error)) runs, and then .finally(...) is executed.

So .finally runs:

Not just “when rejected”.

Not just “when resolved”.

But whenever the promise is resolved or rejected .

Therefore, the correct choice is:

D. When resolved or rejected .

We already know NaN is special: it is not equal to itself.

Check each:

A. value === Number.NaN

Number.NaN is NaN.

NaN === NaN is always false.

This will never be true; cannot reliably detect NaN.

B. value == NaN

NaN == NaN is also always false.

Again, this never detects NaN.

C. isNaN(value)

Global isNaN converts its argument to a number and then checks if the result is NaN.

This can detect NaN, but it may also return true for non-number values that coerce to NaN, such as isNaN( ' foo ' ).

Regardless, it is a standard way to detect if a value is “NaN-like” in JavaScript.

D. Object.is(value, NaN)

Object.is(NaN, NaN) returns true.

This is a strict way to detect a value that is exactly NaN (no coercion).

Therefore, among the given choices, the two viable ways to detect NaN are:

isNaN(value)

Object.is(value, NaN)

We must track synchronous code, setTimeout callbacks (macrotasks), and Promise rejection handling (microtasks).

Step-by-step:

Synchronous code first:

Line 01–03: Schedules a timeout at 1100 ms: logs 1 (later).

Line 04: console.log(2); → logs 2.

Lines 05–09: Construct a new Promise.

The executor runs immediately.

Inside it, another setTimeout is set for 1000 ms:

setTimeout(() = > {

reject(console.log(3));

}, 1000);

Line 09–11: .catch(() = > { console.log(4); }) attached to the promise.

Line 12: console.log(5); → logs 5.

So after all synchronous code, the console has:

2

5

At 1000 ms: inner setTimeout fires

The callback:

() = > {

reject(console.log(3));

}

Inside:

console.log(3) runs first , logging 3.

console.log(3) returns undefined.

Then reject(undefined) is called.

So at about 1000 ms, we log:

3

When the promise is rejected:

The .catch handler is scheduled as a microtask .

After the current macrotask (the timeout callback) completes, the microtask queue runs.

Thus .catch(() = > { console.log(4); }) runs shortly after, in the same 1000 ms tick.

So immediately after 3, the catch handler logs:

4

Now the logs in order are:

2

5

3

4

At 1100 ms: outer setTimeout fires

The callback from line 01 runs:

console.log(1);

Logs: 1.

Final log order:

2 (line 4, sync)

5 (line 12, sync)

3 (1000 ms timeout, then inside logs before reject)

4 (promise catch microtask after rejection)

1 (1100 ms timeout)

Concatenated: 25341.

Therefore, the correct option is:

Answer: A

Study Guide / Concept References (no links):

Event loop: call stack, macrotask queue (timers), microtask queue (promises)

setTimeout scheduling and ordering

Promise rejection, .catch, and microtasks

Evaluation order of function arguments (reject(console.log(3)))

We have a ternary:

account.style.display = accountName.includes(parsedSearchString)

? /* if true */

/* if false */;

Requirement:

If the account matches the search string → show it.

If it does not match → hide it.

For display:

Show: ' block ' (or similar visible value).

Hide: ' none ' .

So we want:

account.style.display = accountName.includes(parsedSearchString)

? ' block '

' none ' ;

That corresponds to option A.

Why others are wrong:

B, C use values suitable for visibility, not display.

D ( ' none ' : ' block ' ) inverts the logic, hiding matches and showing non-matches.

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 References (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.

==================================================

The correct answer is D , because JavaScript executes synchronous code first, while the code after await runs later after the Promise resolves.

The execution order is:

console.log(2);

This runs first, so JavaScript logs:

2

Then this line runs:

asyncCall();

The asyncCall() function starts executing. Inside it, JavaScript reaches:

await delay(1000);

The delay(1000) function returns a Promise that resolves after 1000 milliseconds:

return new Promise(resolve = > {

setTimeout(resolve, delay);

});

Because await pauses the rest of the asyncCall() function, this line does not run immediately:

console.log(1);

Instead, JavaScript continues executing the remaining synchronous code outside the async function:

console.log(3);

So JavaScript logs:

3

After approximately 1000 milliseconds, the Promise returned by delay(1000) resolves. Then the paused async function continues and runs:

console.log(1);

So JavaScript logs:

1

Final console output:

2

3

1

Important JavaScript concepts involved:

An async function always returns a Promise.

The await keyword pauses execution only inside the async function where it appears.

Code outside the async function continues running normally.

setTimeout() schedules its callback to run later, after the current synchronous code has finished.

Therefore, the verified answer is D. 2 3 1 .

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.

References / 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

The correct answers are A, B, and C .

The original variable name appears to contain a typing error. The variable should be used consistently as sampleText:

let sampleText = " The quick brown fox jumps " ;

A is correct because includes() checks whether a string contains a specific substring and returns a Boolean value:

sampleText.includes( ' fox ' );

Since " fox " exists inside " The quick brown fox jumps " , this returns:

true

B is correct after correcting the typing error. The correct expression is:

sampleText.indexOf( ' quick ' ) > -1;

indexOf() returns the index position where the substring is found. If the substring is not found, it returns -1.

Since " quick " exists in the string, sampleText.indexOf( ' quick ' ) returns a value greater than -1, so the expression returns:

true

C is correct after correcting the missing quotation marks and adding a Boolean comparison:

sampleText.indexOf( ' fox ' ) !== -1;

The substring " fox " exists in the string, so indexOf( ' fox ' ) does not return -1. Therefore, the expression returns:

true

D is incorrect because the method name is typed incorrectly. JavaScript string has includes(), not include().

Incorrect:

sampleText.include( ' fox ' )

Correct:

sampleText.includes( ' fox ' )

E is incorrect because JavaScript string matching is case-sensitive:

sampleText.indexOf( ' Quick ' ) !== -1;

The string contains " quick " with a lowercase q, not " Quick " with an uppercase Q, so this returns:

false

Therefore, the verified answers are A, B, and C .

setInterval(fn, delay, arg1, arg2, ...) calls fn every delay ms, passing extra arguments.

To call isDeg every 60 seconds with " cat " :

setInterval(isDeg, 60000, ' cat ' );

Why others are wrong:

A/D: isDeg( ' cat ' ) is called immediately, and its return value is scheduled, not the function.

C: setTimeout runs once after 60 seconds, not repeatedly.

Person.proto is being set, but JavaScript uses Person.prototype for the prototype chain, not Person.proto.

Therefore, job is not on Person.prototype, and instances of Person do not have job via prototype.

myFather is created with new Person(), so:

myFather.firstName is ' John ' .

myFather.job is undefined.

Attempting to call myFather.job() results in:

TypeError: myFather.job is not a function

So option A is correct.