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← JEE Advanced 2021 · Paper 2
MathematicsNumerical+2Question 48 of 57

JEE Advanced 2021 · Paper 2, Mathematics question 10

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 6m2+4n2+8m2n26m_2 + 4n_2 + 8m_2n_2 is ____.

Answer and worked solution

Official answer: 5.9

Derivative.
f2′(x)=4900(x−1)49−29400(x−1)48=4900(x−1)48(x−7).f_2'(x)=4900(x-1)^{49}-29400(x-1)^{48}=4900(x-1)^{48}(x-7).

Extrema. (x−1)48≥0(x-1)^{48}\ge0, so the sign changes only at x=7x=7, from − to +. That is a single local minimum: m2=1m_2=1, n2=0n_2=0.

Value. 6(1)+0+0=66(1)+0+0=6.