Let f_ 1 :(0, ∞) arrow ℝ and f_ 2 :(0, ∞) arrow ℝ be defined by f_ 1 (x)=∫_ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, 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 a_ 1

Correct answer: 0 Given f _ 1 (x) = ∫ _ 0 ^ x (t - 1) (t - 2) ^ 2 ...... (t - 2t) ^ 21 dt ⇒ f ^ ' 1(x) = (x - 1) (x - 2) ^ 2 ..... (x - 21) ^ 21 Checking critical points At all odd integers from 1 to 21 f(x) will have an extrema with 1, 5, 9, 13, 17, 21 being points of minima & 3, 7, 11, 15, 19 being points of maxima. So m _ 1 = 6 & n _ 1 = 5 Hence 2 m _ 1 + 3 n _ 1 + m _ 1 n _ 1 = 57

JEE Advanced 2021Paper 2

Exam: JEE Advanced
Year: 2021
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2021 · Paper 2 · Mathematics · Application of Derivatives

Let f_ 1 :(0, ∞) arrow ℝ and f_ 2 :(0, ∞) arrow ℝ be defined by f_...

Let

f_{1}:(0, \infty) \rightarrow \mathbb{R}

and

f_{2}:(0, \infty) \rightarrow \mathbb{R}

be defined by f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, \quad x>0 and

f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, \quad x \gt 0

where, for any positive integer

n

and real numbers

a_{1}, a_{2}, \ldots, a_{n}, \prod_{i=1}^{n} a_{i}

denotes the product of

a_{1}, a_{2}, \ldots, a_{n} .

Let

m_{i}

and

n_{i}

, respectively, denote the number of points of local minima and the number of points of local maxima of function

f_{i}, i=1,2

, in the interval

(0, \infty)

. The value of~

2{m}_{1} + 3{n}_{1} + {m}_{1}{n}_{1}

& is _____. Let f_ 1 :(0, ∞) arrow ℝ and f_ 2 :(0, ∞) arrow ℝ be defined by f_ 1 (x)=∫_ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x>0 and f_ 2

Correct answer

Step-by-step solution

Given

${f}_{1}\left(x\right) = \mathop{∫}_{0}^{x}\left(t - 1\right){\left(t - 2\right)}^{2}......{\left(t - 2t\right)}^{21}dt$

$⇒ {f}^{'}1\left(x\right) = \left(x - 1\right){\left(x - 2\right)}^{2}.....{\left(x - 21\right)}^{21}$

Checking critical points

Let f_ 1 :(0, ∞) arrow ℝ and f_ 2 :(0, ∞) arrow ℝ be defined by f_ 1 (x)=∫_ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x>0 and f_ 2

At all odd integers from $1$ to $21$ $f\left(x\right)$ will have an extrema with $1, 5, 9, 13, 17, 21$ being points of minima & $3, 7, 11, 15, 19$ being points of maxima.

So ${m}_{1} = 6$ & ${n}_{1} = 5$

Hence $2{m}_{1} + 3{n}_{1} + {m}_{1}{n}_{1} = 57$

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