 Hello and welcome to the session. In this session, we will discuss a question which says that Find A, B, C when f of x is equal to A x square plus B x plus C f of 0 is equal to 3, f of 2 is equal to 5, f of 3 is equal to 10 Determine the quadratic function f of x and find its value when x is equal to 1 Now before starting the solution of this question, we should know our result and that is Kramer's rule. Now for the system linear equations x, y, z that is A1 x plus B1 y plus C1 z is equal to D1 A2 x plus B2 y plus C2 z is equal to D2 A3 x plus B3 y plus C3 z is equal to D3 By Kramer's rule we can find x, y and z and here x is equal to determinant dx over determinant d y is equal to determinant dy over determinant d and z is equal to determinant dz over determinant d where determinant d is not equal to 0 and determinant d is equal to the determinant with the elements in first row as A1 B1 C1 elements in the second row as A2 B2 C2 and elements in the third row as A3 B3 C3 and determinant dx can be obtained by replacing the x coefficients that is A1 A2 and A3 by the constants d1 d2 and d3 so the determinant dx will be equal to the determinant with the elements in the first row as D1 B1 C1 elements in the second row as D2 B2 C2 and elements in the third row as D3 B3 C3 and the determinant dy is equal to the determinant with the elements in the first row as A1 D1 C1 elements in the second row as A2 D2 C2 and elements in the third row as A3 D3 C3 that is in this case we will replace the y coefficients with the constants D1 D2 and D3 and the determinant dz is equal to the determinant with the elements in the first row as A1 B1 D1 elements in the second row as A2 B2 D2 and elements in the third row as A3 B3 D3 now this result will work out as a key idea for solving out this question and now we will start with the solution now here we have to find A B and C when f of x f of 0 f of 2 and f of 3 are given to us now given f of x is equal to A x square plus B x plus C therefore f of 0 is equal to A into 0 plus B into 0 plus C that is putting x is equal to 0 in this we can find f of 0 also it is given f of 0 is equal to 3 therefore from these two equations we have A into 0 plus B into 0 plus C is equal to 3 similarly f of 2 will be equal to now let it be equation A so putting x is equal to 2 in A f of 2 will be equal to A into 2 square plus B into 2 plus C and also f of 2 is given as 5 so this implies 4A plus 2B plus C is equal to 5 now f of 3 will be equal to now putting x is equal to 3 in A f of 3 will be equal to A into 3 square plus B into 3 plus C and f of 3 is also given equal to 10 so this implies 9A plus 3B plus C is equal to 10 now let this be equation number 1 this has 2 and this equation has 3 now using the result which is given in the key idea so for equations 1, 2 and 3 we have determinant D is equal to the determinant with the elements in the first room as A1, B1, C1 now here A1 is 0, B1 is 0 and C1 is 1 so this will be the determinant with the elements in the first room as 0, 0, 1 and the elements in the second room as A2 which is 4B2 which is 2 and C2 which is 1 here and the elements in the third room as A3 that is 9B3 that is 3 and C3 which is 1 on solving this will be equal to 0 into 2 minus 3 in the hole minus 0 into 4 minus 9 in the hole plus 1 into 12 minus 18 in the hole so this will be equal to 0 minus 0 plus 1 into 12 minus 18 in the hole will give minus x which is not equal to 0 therefore the solution exists now we will find the determinant dx which is equal to the determinant with the elements in the first room as D1, B1, C1 now here D1 is 3, B2 is 5 and D3 is 10 so using the values from these equations the determinant dx will be equal to the determinant with the elements in the first room as 301 elements in the second room as 5, 2, 1 and elements in the third room as 10, 3, 1 on solving this is equal to 3 into 2 minus 3 the hole minus 0 into 5 minus 10 the hole plus 1 into 5 into 3 is 15 minus 2 into 10 is 20 the hole so this is equal to 3 into 2 minus 3 is minus 1 so 3 minus 1 is minus 3, 0 into anything is 0 plus 1 into 15 minus 20 the hole and 15 minus 20 is minus 5 and minus 5 into 1 is minus 5 so this will be equal to minus 8 now the determinant dy is equal to the determinant with the elements in the first room as 031 elements in the second room as 451 elements in the third room as 9, 10, 1 so this will be equal to 0 into 5 minus 10 the hole minus 3 into 4 minus 9 the hole plus 1 into 4 into 10 is 40 minus 9 into 5 is 45 the hole now this is equal to now 0 into anything is 0 and minus 3 into 4 minus 9 the hole now 4 minus 9 the hole is minus 5 and minus 5 into minus 3 is plus 15 here it will be plus 1 into 40 minus 45 the hole now 40 minus 45 is minus 5 and minus 5 into plus 1 is minus 5 so this is equal to 10 now the determinant dz is equal to the determinant with the elements in the first room as 003 elements in the second room as 4, 2, 5 and elements in the third room as 9, 3, 10 so this is equal to 0 into 2 into 10 is 20 minus 5 into 3 is 15 the hole minus 0 into 4 into 10 is 40 minus 9 into 5 is 45 the hole plus 3 into 4 into 3 is 12 minus 9 into 2 is 18 the hole now this is equal to 0 minus 0 plus 3 into minus 6 which is equal to minus 18 now it will be equal to the determinant dx over determinant d now the determinant d is equal to minus 6 and the determinant dx is equal to minus 8 so this implies a is equal to minus 8 over minus 6 which is equal to 4 by 3 now b is equal to determinant dy over determinant t now the determinant dy is equal to 10 so this is equal to 10 over minus 6 so this implies b is equal to minus 5 by 3 equal to determinant dz over determinant t now the determinant dz is equal to minus 18 so c will be equal to minus 18 over minus 6 which is equal to 3 now given x is equal to ax square plus x plus c so putting the values of a v and c this implies f of x is equal to 4 by 3 x square plus or minus 5 by 3 will be equal to minus 5 by 3 into x plus 3 also we have to determine the value of f of x when x is equal to 1 now this is the quadratic function f of x so when x is equal to 1 then f of 1 will be equal to now replacing x is equal to 1 in this equation it will be 4 by 3 into 1 square minus 5 by 3 into 1 plus 3 which is equal to 4 by 3 minus 5 by 3 plus 3 which is equal to 4 minus 5 plus 9 whole upon 3 which is equal to 8 by 3 therefore we have got a is equal to 4 by 3 b is equal to minus 5 by 3 and c is equal to 3 and f of 1 that is the value of the quadratic function when x is equal to 1 is equal to 8 by 3 so this is the solution of the given question and that's all for this session hope you all have enjoyed the session