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Maths Question 2 – JEE-MAIN 2026

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Consider the quadratic equation (n2−2n+2)x2−3x+(n2−2n+2)2=0,n∈R. Let α be the minimum value of the product of its roots and β 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 α and the common ratio is αβ, is:

Identify the coefficients of the quadratic equation and express them in terms of a single variable to simplify the problem.

● Video Walkthrough
Step 1: Analyze the quadratic equation and define key terms✦ Active

The given quadratic equation is (n2−2n+2)x2−3x+(n2−2n+2)2=0. Let k=n2−2n+2. This can be rewritten as k=(n−1)2+1. Since (n−1)2≥0, the minimum value of k is 1 (when n=1). Thus, k≥1. The equation becomes kx2−3x+k2=0. The product of roots is P=k2k=k. The sum of roots is S=−(−3)k=3k.

Step 2: Determine α and β○ Expand

α is the minimum value of the product of roots, P. Since P=k and k≥1, the minimum value of P is 1. So, α=1. β is the maximum value of the sum of roots, S. Since S=3k and k≥1, to maximize S, k must be minimized. The minimum value of k is 1. So, the maximum value of S is 31=3. Thus, β=3.

Step 3: Calculate the sum of the first six terms of the G.P.○ Expand

The first term of the G.P. is a=α=1. The common ratio is r=αβ=13. The sum of the first six terms of a G.P. is S6=a1−r61−r. Substituting the values:

S6=1⋅1−(13)61−13 S6=1−172923 S6=729−172923 S6=72872923 S6=728729×32 S6=364243
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