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← JEE Advanced 2021 · Paper 2
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JEE Advanced 2021 · Paper 2, Mathematics question 9

Passage

Question Stem for Question Nos. 9 and 10

Let f1:(0,∞)→Rf_1: (0, \infty) \to \mathbb R and f2:(0,∞)→Rf_2: (0, \infty) \to \mathbb R be defined by
f1(x)=∫0x∏j=121(t−j)j dt,x>0f_1(x) = \int_0^x \prod_{j=1}^{21}(t - j)^j\,dt,\qquad x > 0
and
f2(x)=98(x−1)50−600(x−1)49+2450,x>0,f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450,\qquad x > 0,
where, for any positive integer nn and real numbers a1,a2,…,ana_1, a_2, \ldots, a_n, ∏i=1nai\prod_{i=1}^n a_i denotes the product of a1,a2,…,ana_1, a_2, \ldots, a_n. Let mim_i and nin_i, respectively, denote the number of points of local minima and the number of points of local maxima of function fif_i, i=1,2i = 1, 2, in the interval (0,∞)(0, \infty).

The value of 2m1+3n1+m1n12m_1 + 3n_1 + m_1n_1 is ____.

Answer and worked solution

Official answer: 56.9

Derivative. f1′(x)=∏j=121(x−j)jf_1'(x)=\prod_{j=1}^{21}(x-j)^j. Its sign changes only at roots of odd multiplicity, j=1,3,…,21j=1,3,\dots,21.

Sign pattern. On (0,1)(0,1) the sign is (−1)1+2+⋯+21=(−1)231<0(-1)^{1+2+\cdots+21}=(-1)^{231}<0. It then flips at each odd root.

  • Minima (− to +) at 1,5,9,13,17,211,5,9,13,17,21: m1=6m_1=6.
  • Maxima (+ to −) at 3,7,11,15,193,7,11,15,19: n1=5n_1=5.

Value. 2(6)+3(5)+6⋅5=572(6)+3(5)+6\cdot5=57.