Here, f: A—>B

f(x)= (x-1)/(x+2)  ;  x   -2 

A= {-1,0,1,2,3,4}

B= {-2,1,-1/2,0,1/2,1/4,2/5}

Range = {-2,-1/2,0,1/4,1/2,2/5}

As range is not equal to codomain so the given function is not bijective. 

We can make it bijective by omitting {1} from set B 

Program to input any number and display number of odd numbers in it;

#include<stdio.h>

#include<stdlib.h>

#include<conio.h>

#include<string.h>

int main() {

int i, number, num1, num2=0, num3, num4 =0, rem, rem1, rem2, rev = 0;

printf("Enter your number ==> ");

scanf("%d", &number);

num1 = number;

while(num1 != 0) {

rem = num1%10;

rem1 = rem%2;

if (rem1 != 0) {

num2 = num2*10 + rem;

}

num1 /= 10;

}

num3 = num2;

while (num2 != 0) {

...

The answer is A 

In the depletion layer, electrons combine with holes producing charge density so that region is depleted of charge carriers  but it has charge density .

Conclusion: Neutral but still having non zero charge density .

This picture develops a clear understanding.