JEE Advance - Mathematics (2021 - Paper 2 Online - No. 9)
Let f1 : (0, $$\infty$$) $$\to$$ R and f2 : (0, $$\infty$$) $$\to$$ R be defined by $${f_1}(x) = \int\limits_0^x {\prod\limits_{j = 1}^{21} {{{(t - j)}^j}dt} } $$, 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, $$\prod\nolimits_{i = 1}^n {{a_i}} $$ 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 $$2{m_1} + 3{n_1} + {m_1}{n_1}$$ is ___________.
The value of $$2{m_1} + 3{n_1} + {m_1}{n_1}$$ is ___________.
Answer
57.00
Explanation
$${f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt} $$
$$ \Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$$
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, $$2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$$
$$ \Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$$
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, $$2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$$
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