JEE Advanced 2021 Paper 2 · Q10 · Maxima & Minima

Correct answer:6

Solution

$f_2'(x) = 98\cdot 50(x-1)^{49} - 600\cdot 49(x-1)^{48} = 49\cdot 100(x-1)^{48}(x-1-6) = 4900(x-1)^{48}(x-7)$. Since $(x-1)^{48}\ge 0$, $f_2'$ changes sign only at $x=7$, going from negative to positive. So $m_2 = 1$ (local min at $x=7$), $n_2 = 0$. Thus $6m_2 + 4n_2 + 8m_2 n_2 = 6 + 0 + 0 = 6$.

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