Take a look at the following code:

1    let x = 1;
2    function f1()
3    {
4        let x = 2;
5        console.log(x);
<!--  here we have reassigned the value of variable x into 2 so the console.log will print 2 -->
6    }
7    console.log(x);
<!-- here variable x will take the value of 1 as x is declared as a global variable -->

Explain why line 4 and line 6 output different numbers.

Take a look at the following code:

let x = 10

function f1()
{
    console.log(x)
    let y = 20
}

console.log(f1())
console.log(y)

What will be the output of this code. Explain your answer in 50 words or less. Answer - 10 : because line 26 is accessing the variable declared on line 1 due to closure. undefined : because the f1 function isn't returning anything. reference error: because the y variable only exists in the scope of the f1 function.

Take a look at the following code:

const x = 9;

function f1(val) {
  val = val + 1;
  return val;
}

f1(x);
console.log(x);

const y = { x: 9 };

function f2(val) {
  val.x = val.x + 1;
  return val;
}

f2(y);
console.log(y);

What will be the output of this code. Explain your answer in 50 words or less. 9 { x: 10 }

line 58 & 68 print anything because the function call is not wrapped in console.log(). line 59 prints 9 because x is accessible here within the global scope. line 69 will print { x: 10 } because y has been copied with it's reference, not value.