Verified Solution • 2025–26 Session moderate

QJEEAdv21Ma79ed: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma79ed Solution: Let f 1 : (0, ) R and f 2 : (0, ) R be defined by , x > 0 and , …

Let f1 : (0, ∞\infty ) →\to R and f2 : (0, ∞\infty ) →\to R be defined by f1(x)=∫0x∏j=121(t−j)jdt{f_1}(x) = \int\limits_0^x {\prod\limits_{j = 1}^{21} {{{(t - j)}^j}dt} } , x > 0 and {f_2}(x) = 98{(x - 1)^{50}} - 600{(x - 1)^{49}} + 2450,x > 0 , where, for any positive integer n and real numbers a1, a2, ....., an, ∏i=1nai\prod\nolimits_{i = 1}^n {{a_i}} denotes the product of a1, a2, ....., an. Let mi and ni, respectively, denote the number of points of local minima and the number of points of local maxima of function fi, i = 1, 2 in the interval (0, ∞\infty ).

The value of 2m1+3n1+m1n12{m_1} + 3{n_1} + {m_1}{n_1} is ___________.

Step-by-Step Verified Solution

f1(x)=∫0x(t−1)1(t−2)2(t−3)3(t−4)4....(t−21)21dt{f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt}

⇒f1′(x)=(x−1)(x−2)2(x−3)3(x−4)4....(x−21)21\Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}

Sign Scheme for f1'(x)

From sign scheme of f1'(x), we observe that f(x) has local minima at x = 4k + 1, k ∈\in W i.e. f1'(x) changes sign from −- ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f(x) has local maxima at x = 4k + 3, k ∈\in W i.e. f1'(x) changes sign from + ve to −- ve, which are x = 3, 7, 11, 15, 19.

So, m1 = number of local minima points = 6

and n1 = number of local maxima points = 5

Hence, 2m1+3n1+m1n1=2×6+3×5+6×5=572{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57

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