What JavaScript-Developer-I Questions and Answers feature?

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.

objA is { value: 10 }.

objB = objA; → both variables reference the same object .

changeValue(objB);:

obj.value = obj.value / 2; // modifies the shared object

So:

objA.value becomes 10 / 2 = 5.

Thus result is 5.

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