JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q2

← JEE Main 2026 · 6 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 2 of 75

Consider the quadratic equation $(n^2-2n+2)x^2-3x+(n^2-2n+2)^2=0$ , $n\in\mathbf R$ . Let $\alpha$ be the minimum value of the product of its roots and $\beta$ 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 $\alpha$ and the common ratio is $\dfrac\alpha\beta$ , is :

\dfrac{61}{37}

\dfrac{121}{81}

\dfrac{364}{243}

\dfrac{1093}{729}

Answer and worked solution

Official answer: C

Substitute $t=n^2-2n+2=(n-1)^2+1\ge1$ .

Product of roots: $\dfrac{t^2}t=t$ , minimum $\alpha=1$ .
Sum of roots: $\dfrac3t$ , maximum $\beta=3$ .

G.P. First term 1, ratio $\tfrac13$ :
$S_6=\frac{1-3^{-6}}{2/3}=\frac32\cdot\frac{728}{729}=\frac{364}{243}.$

(C)

Topics:Roots and coefficients Geometric progression

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JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 2