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JEE Advanced 2021 Paper 2

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Let

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

and

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

be defined by

f_{1}(x)=\int\limits_{0}^{x} \prod_{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

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)

Section: Mathematics

Question : 47 of 57

Marks: +1, -0

The value of

2 m_{1}+3 n_{1}+m_{1} n_{1}

Solution:

f_{1}(x)=\int\limits_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t

f_{1}'(x)=\prod_{j=1}^{21}(x-j)^{j}=(x-1)(x-2)^{2}(x-3)^{3}\ldots (x-21)^{21}

Checking the sign scheme of

f_{1}'(x)

x=1,2,3,\ldots ,21

We get

f_{1}(x)

has local minima at

x=1,5,9,13,17,21

and local maxima at

3,7,11,15,19

\Rightarrow m_{1}=6, n_{1}=5

So,

2 m_{1}+3 n_{1}+m_{1} n_{1}

=2 \times 6+3 \times 5+6 \times 5=57

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