 Hi, and welcome to the session. I am Deepika here. Let's discuss a question. Find the least value of A such that the function f given by A affects is equal to x square plus Ax plus 1 is strictly increasing on an open interval 1 to 2. Now in this question, we will find the value of A such that the function f given by fx is equal to x square plus Ax plus 1 is strictly increasing on the open interval 1 to 2. Let's start the solution. Given fx is equal to f plus Ax plus 1, therefore, this is equal to 2x plus A using function, function on an open interval 1, 2. Therefore, this is greater than 0 and less than x less than 2. Now, 1 less than x, x less than 2, this implies 2 less than 2x less than 4. And this implies 2 plus A less than 2x plus A less than 4 plus A. And this implies 2 plus A less than, now this is our f dash x, less than 4 plus A. If fx is strictly increasing function on an interval, should be greater than 0 for a strictly increasing function fx, f dash x is greater than 2 plus A. This implies should be greater than or equal to 0. And this implies A is greater than or equal to minus 2. Hence, the least value of A is minus 2. The least value of A, such that the function f given by fx is equal to x square plus Ax plus 1 is strictly increasing on an open interval 1 to 2, is A is equal to minus 2. Hence, the answer for the above question is A is equal to minus 2. I hope the question is clear to you. Bye and take care.