 Hello, welcome to the session. In this session, first of all we will discuss the method of intervals for solving quadratic inequality. Now suppose we have to solve the inequality x minus x1 the whole into x minus x2 the whole into x minus x3 the whole and so on up to x minus xn minus 1 the whole into x minus xn the whole is less than 0 where x1, x2, x3 and so on up to xn are distinct real numbers. Then first of all we will assume that is greater than xn minus 1 is greater than and so on is greater than x3 is greater than x2 is greater than x1. Now we will get these points on the real number line and consider a polynomial f of x is equal to x minus x1 the whole into x minus x2 the whole into x minus x3 the whole and so on x minus xn minus 1 the whole into x minus xn the whole. Now here we have drawn a number line in which we have plotted these points. Now in this number line we will start with the positive sign from the right towards the left also from the number line you can see that for all x greater than xn expressions are positive. Hence for x greater than xn we have greater than 0 that is f of x is positive. Now let us discuss the second case when x is lying between xn minus 1 and xn and for this case this expressions except x minus xn will be positive this means f of x will be less than 0 that is for this case f of x is negative which is denoted by this interval that is if x is lying between xn minus 1 and xn then f of x will be less than 0. Similarly we can discuss it by taking x in some other intervals. Now let us recall the method of intervals and this in the first step on the number line the numbers x1, x2 and so on up to xn are arranged in the increasing order of the magnitude then we will put the positive sign in the interval to the right of the largest number. Now here the largest number is xn so we are putting the positive sign on the right of the largest number then in the next interval that is from right to left we will place a negative sign and then again the positive sign and then again the negative sign and so on. Now the solution of the inequality that is personal equality that is f of x will be less than 0 in the intervals with negative sign and f of x will be greater than 0 in the intervals with positive sign. Now let us discuss this method with the help of an example. Let us take the inequality x minus 1 the whole into x minus 2 the whole into x minus 3 the whole into x minus 4 the whole is less than 0. Now let f of x is equal to x minus 1 the whole into x minus 2 the whole into x minus 3 the whole into x minus 4 the whole. Now let us plot these points on the number line so we have plotted these points on the number line in the order of the increasing magnitude. Now here you can see five regions first is from form to infinity second is from three to four and fourth region from one to two and the fifth region from one to minus infinity of the greatest number and here the greatest number is four positive sign here then a negative sign towards the left and then positive again then negative and then again positive in the region with positive signs greater than 0 greater than 0 when x belongs to open interval over to infinity union union the open interval 1 to minus infinity will be less than 0 in the region with negative sign that this f of x is less than 0 when x belongs to the open interval union the open interval 1 to 2 maximum and minimum values. Now we know that the graph is equal to a x square plus b x plus c is a truth that is contained upwards when e is greater than 0 and it is contained downwards minimum values we have to find the limits within which the value of a x square plus b x plus c will lie for real values of x with the help of the graphs. Now for the part 1 when a is greater than 0 then we have drawn the graph of the quadratic function now here a b is the minimum value of the function which occurs at the vertex so we can write that the minimum value a b is equal to 4ac minus b square whole upon 4a at the vertex there the value of x is equal to minus b over 2a that is the distance of a. Now in the second case when e is less than 0 then we have drawn the graph of the quadratic function now here the maximum value of the function that is the maximum value a b is equal to 4ac minus b square whole upon 4a which occurs at the vertex where the value of x is equal to minus b upon 2a that is the distance of b because theoretically f of x will be maximum or minimum the vertex that is where the value of x is minus b upon 2a will be equal to a into minus b upon 2a whole square plus b into minus b over 2a whole plus c which is further equal to a b square upon 4a square minus b square upon 2a plus c and on solving this will give 4ac minus b square whole upon 4a for when a is greater than 0 the minimum value of the function f of x is 4ac minus b square over 4a and the maximum value of the function in the case of a is less than 0 is now the quadratic function f of x plus bx plus c over px square plus qx plus a bc pqmr are the constants you are not equal to 0 set with the help of an example the value of 2x square minus 2x plus 4 cannot lie between minus 7 and 1 now let us start with its solution let y is equal to 2x square minus 2x plus 4 over 1x square minus 4x plus 3 now by possibility this implies minus 4x plus 3 whole is equal to and further on solving this implies y minus 2 the whole into x square minus 2 into 2 y minus 1 the whole into x plus 3 y minus 4 minus 4 the whole is equal to 0 now this is the quadratic equation in x therefore the determinant d will be greater than equal to 0 this implies b square minus 4ac is greater than equal to 0 and the whole this whole is 4 into y minus 2 the whole into 3 y minus 4 the whole is greater than equal to 0 which implies 4 into 4 y square plus 1 minus 4 y the whole minus 4 into 3 y square minus 10 y plus 8 the whole is greater than equal to 0 now dividing 3 out by 4 and further solving it gives 4 y square plus 1 minus 4 y minus 3 y square plus 10 is greater than equal to 0 and solving this implies y square plus 6 y minus 7 is greater than equal to 0 which further gives y plus 1 the whole is greater than equal to 0 this inequality by the method of intervals which we have discussed earlier and we have given the intervals by using the method of intervals so let y plus 7 the whole into y minus 1 the whole is f of x will be greater than equal to 0 in the region is in the intervals the solution of the given inequality open interval minus infinity to minus 7 union that is within this range x will be greater than equal to 0 now we have already learnt how to solve the inequalities of the type x minus a the whole into x minus b the whole into x minus greater than 0 where a, b, c, t are the constants now let us know how to solve the inequality is 2 is greater than 4 x minus 5 the range of the variable x and this inequality will take us we cannot multiply the whole into 6 x minus 3 a whole is a positive quantity we will solve this with another method for this we will take all the terms of the given inequality on the left side that is x minus 2 whole upon x plus 2 the whole by whole upon 6 x minus 2 the whole is greater than 0 of the denominators this x plus 2 the whole into 6 x minus 3 in the numerator it will be into 6 x minus 3 the whole minus we get 2 x square minus 18 x plus 16 whole into 6 x minus 3 the whole is greater than 0 now taking 2 common from the numerator from the factor 6 x minus 3 in the denominator we get 2 into x square minus 9 x plus 8 the whole whole upon the whole into 6 into x minus that is 1 by 2 the whole is greater than 0 now on solving this implies x square minus 9 x into x plus 2 the whole into x minus 1 by 2 the whole is greater than 0 now multiplying throughout by 3 and then actualizing the numerator we get minus 1 the whole into x minus 8 the whole all upon x plus 2 the whole into x minus 1 by 2 the whole is greater than 0 now let this equation be a so now multiplying both sides x plus 2 whole square into x minus 1 by 2 whole square which is a positive quantity for the S under consideration we get 1 the whole into x minus 8 the whole into x plus 2 whole square into x minus 1 by 2 whole square all upon x plus 2 the whole into x minus 1 by 2 the whole is greater than 0 1 the whole into x minus 8 the whole into x plus 2 the whole into x minus 1 by 2 the whole is greater than 0 now this inequality can be solved by the method of interiors which we have discussed earlier from these 2 inequalities that the solution is a the whole into x minus b the whole all upon x plus 3 the whole into x minus d the whole is greater than 0 x minus a the whole into x minus b the whole into x minus c the whole into x minus d the whole is greater than 0 so you have learnt about method of interiors minimum values So this is the Fleece Session. Hope you all have enjoyed the session.