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JEE Advanced 2021 Paper 2

Paragraph for next 2 questions

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 : 48 of 57

Marks: +1, -0

The value of

6 m_{2}+4 n_{2}+8 m_{2} n_{2}

Solution:

f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450

\Rightarrow \;\; f_2'(x)=2 \times 49 \times 50 (x-1)^{49}-50 \times 12 \times 49 (x-1)^{48}

=50 \times 49 \times 2 (x-1)^{48}(x-1-6)

=50 \times 49 \times 2 (x-1)^{48}(x-7)

f_{2}(x)

has local minimum at

x=7

and no local minimum

x=7

and no local minimum

\Rightarrow \;\; m_{2}=1, n_{2}=0

6 m_{2}+4 n_{2}+8 m_{2} n_{2}

=6 \times 1+4 \times 0+8 \times 1=6

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