Consider the quadratic equation (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0, n R . Let be the minimum value of the product of its roots and be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is and the common ratio is , is :
C. 364 243
The given quadratic equation is (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0. Let k = n^2 - 2n + 2 = (n-1)^2 + 1. Since (n-1)^2 0 for all n R , the minimum value of k is 1 at n=1. The product of the roots is given by P = (n^2 - 2n + 2)^2 n^2 - 2n + 2 = n^2 - 2n + 2 = k. The minimum value of the product of the roots is = (k) = 1. The sum of the roots is given by S = 3 n^2 - 2n + 2 = 3 k . The maximum value of the sum of the roots occurs when k is minimum. Thus, = ( 3 k ) = 3 1 = 3. We are given a Geometric Progression (G.P.) whose first term is a = = 1 and common ratio is r = = 1 3 . The sum of the first six terms of this G.P. is: S_6 = a 1 - r^6 1 - r = 1 1 - ( 1 3 )^6 1 - 1 3 S_6 = 1 - 1 729 2 3 = 728 729 2 3 S_6 = 728 729 3 2 = 364 243 Answer: 364 243
Related: Mathematics — Quadratic Equation · All PYQ Banks