 Hello and welcome to the session. In this session we will discuss the question which says that a function f is defined on the set of integers as follows. fx is equal to 2x plus 3 when x is less than equal to 0 and 3 into x plus 1 the whole when x is greater than 0. Then first part is find the domain of the function second part is find the range of the function and the third part is find the value of f of minus 1 f of 0 f of 1 also finds f of minus 1 by 2 plus 2 into f of 4 by 3 minus after the solution of this question we shall know our result. And that is function defined of integers which is denoted by z function will be equal to which belongs to the set of integers such that that is will be equal to the set containing such that x belongs of this function. Now this result will work out as a key idea and now we will start with the solution. Now where the function is given to us is 3 when x is less than equal to 0 in the whole when it finds one function even when it involves you have to find the domain of now using this result which is given in the key idea now in this list for x to be integer the function that is f of x is equal to 2x plus 3 when x is less than equal to 0 f of x is equal to 2x plus 3 which is equal to 2 into 0 plus 3 which is equal to 3. Now for x is equal to minus 1 to 2x plus 3 which is equal to 2 into minus 1 plus 3 which is equal to minus 2 plus 3 which is 1 when x is greater than 0 will be equal to just 1 the whole x will be equal to 3 into x plus 1 the whole which is equal to 3 into 1 plus 1 the whole which is equal to 3 into 1 which is 1 is 2 so 3 into 2 is 3 into 2 plus 1 the whole whole which is equal to 3 into 2 plus 1 is 3 and 3 into 3 is, we have discussed some of the cases in Deja, when f of x is defined that is f of x is also an integer. So from this we can conclude that when sum of integers equal to 0 and x is greater than 0 then f of x is defined that is f of x is also an integer. For the integers the function f that is the function f of x is also an integer which is given in the key idea. So the range continuous of f of x when x belongs to the domain. Now x is an integer and for different values of x we will get therefore the range of this function the set of all integers 0 will never have now here minus 1 that is x is less than equal to 0 equal to 0 then f of x is equal to 2 x plus then we can write when equal to 0 minus 1 is equal to 2 into minus 1 plus 3 which is equal to minus 2 plus 3 which is equal to 1 to 0 here. So when x is less than equal to 0 then f of x is equal to 2 x plus 3 therefore f of 0 will be equal to 2 into 0 plus 3 which is equal to 3. Now next we have now here x is equal to 1 which is greater than 0. Now when x is greater than 0 then f of x is equal to 3 into x plus 1 the whole then f of x is equal to 3 into x plus 1 the whole therefore f of 1 is equal to 3 into 1 plus 1 the whole which is equal to 3 into 1 plus 1 is 2 and 3 into 2 will be equal to minus 1 by 2 minus 1 by 2. Now here x is equal to minus 1 by 2 which is less than 0 and for x is less than equal to 0 is equal to 2 x plus 1 by 2 is equal to 2 into minus 1 by 2. Now here 2 will be cancelled with 2 so it will be minus 1 plus 3 which is equal to 2 is greater than 0 therefore greater than 0 we have into x plus 1 the whole therefore is equal to 3 into 4 by 3 plus 1 the whole which will be equal to 3 into on taking the lc here it will be 2 plus 3 whole upon 3 this will be equal to f of 0 is equal to minus 1 by 2 plus on putting these values here this will be 2 plus just further equal to 2 plus 14 minus 3 solving is equal to so this is the solution of the given question and that's all for this session hope you all have enjoyed the session.