Mathematics · 2026
JEE Main · 28 January 2026, Shift 2 · Q4
Let the arithmetic mean of 1/a and 1/b be 5/16, a >2. If α is such that a, 4, α, b are in A.P., then the equation α x^2- a x+2(α-2 b )=0 has:
Let the arithmetic mean of $\displaystyle \frac{1}{\mathrm{a}}$ and $\displaystyle \frac{1}{\mathrm{~b}}$ be $\displaystyle \frac{5}{16}, \mathrm{a}>2$. If $\displaystyle \alpha$ is such that $\displaystyle \mathrm{a}, 4, \alpha, \mathrm{~b}$ are in A.P., then the equation $\displaystyle \alpha x^2-\mathrm{a} x+2(\alpha-2 \mathrm{~b})=0$ has :
Official answer
From NTA’s final answer key for this paper.
(4)
one root in $\displaystyle (1,4)$ and another in $\displaystyle (-2, 0)$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.