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_

Correct answer: 0 f _ 2 (x) = 98(x - 1)50 - 600(x - 1)49 + 2450, x > 0 f _ 2 ^ ' (x) = 98 × 50(x - 1)49 - 600 × 49 (x - 1)48 = 4900(x - 1)48((x - 1) - 6) = 4900 (x - 1)48 (x - 7) So extrema is at x = 7 only, which is minima m _ 2 = 1, n _ 2 = 0 Hence 6 m _ 2 + 4 n _ 2 + 8 m _ 2 n _ 2 = 6

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 > 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

6{m}_{2} + 4{n}_{2} + 8{m}_{2}{n}_{2}

& is ____.

Correct answer

Step-by-step solution

{f}_{2}\left(x\right) = 98\left(x - 1\right)50 - 600\left(x - 1\right)49 + 2450, x > 0 $$ {f}_{2}^{'}\left(x\right) = 98 \times 50\left(x - 1\right)49 - 600 \times 49 \left(x - 1\right)48 $$ = 4900\left(x - 1\right)48\left(\left(x - 1\right) $- 6\right)$

$= 4900 \left(x - 1\right)48 \left(x - 7\right)So extrema is at$ x = 7 $only, which is minima$ {m}_{2} = 1, {n}_{2} = 0 $Hence$ 6{m}_{2} + 4{n}_{2} + 8{m}_{2}{n}_{2} = 6 $$

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