 Hello and welcome to the session. Let us understand the following problem today. Let a is equal to 1 sin theta 1 minus sin theta 1 sin theta minus 1 minus sin theta 1 where 0 is less than equal to theta is less than equal to 2 pi then a determinant a is equal to 0 b determinant a belongs to open interval 2 comma infinity c the determinant a belongs to open interval 2 comma 4 and d the determinant a belongs to close interval 2 comma 4. Now let us write the solution. Given to us is a is equal to 1 sin theta 1 minus sin theta 1 sin theta minus 1 minus sin theta 1. Now applying r 1 tends to r 1 plus r 3. Applying row pressure on this row so we get 1 minus 1 0 sin theta minus sin theta 0 1 plus 1 2 plus 2 rows as it is minus sin theta 1 sin theta minus 1 minus sin theta 1. Now value of determinant a is equal to that is determinant of a which is 2 multiplied by now expanding through this so eliminating this column and row we get sin square theta plus 1. Therefore maximum of determinant of a is equal to 4 minimum of determinant of a is equal to 2 1 is less than equal to sin theta is less than equal to 1. Here we can see that if sin square theta is equal to 1 then value of determinant a will be 4 so maximum is 4 and if the value of sin square theta is 0 then the value of determinant a will be 2 so minimum of determinant a is equal to 2 as sin theta lies between 1 and minus 1. So determinant of a belongs to the minimum is 2 close intervals 2 comma maximum is 4 2 comma 4. So the correct option is d. I hope you understood the problem bye and have a nice day.