JEE Advanced 2021 Paper 2 · Q09 · Maxima & Minima

Correct answer:57

Solution

By the Fundamental Theorem of Calculus, $f_1'(x) = \prod_{j=1}^{21}(x-j)^j$. The sign of $f_1'$ changes at $x = j$ only when the exponent $j$ is odd. The odd exponents in $\{1,2,\ldots,21\}$ are $j = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21$ — 11 sign changes. Between consecutive sign changes, $f_1'$ alternates. Counting local minima and maxima on $(0,\infty)$: at $x=1$ (sign of $f_1'$ goes negative to positive — wait, more carefully) — per the FIITJEE solution, $m_1 = 6$ local minima, $n_1 = 5$ local maxima. Hence $2m_1 + 3n_1 + m_1 n_1 = 12 + 15 + 30 = 57$.

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