QJEEAdv21Ma1a57: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma1a57 Solution: Let f 1 : (0, ) R and f 2 : (0, ) R be defined by , x > 0 and , …

${f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt}$

$\Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$

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, $2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$

Also, ${f_2}(x) = 98{(x - 1)^{50}} - 600{(x - 1)^{49}} + 2450$

$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{49}} - 600 \times 49{(x - 1)^{48}}$

$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 1 - 6)$

$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 7)$


Clearly, m₂ = 1 and n₂ = 0

So, $6{m_2} + 4{n_2} + 8{m_2}{n_2} = 6 + 0 + 0 = 6$