Let $f_{1}:(0, \infty) \rightarrow \mathbb{R}$ and $f_{2}:(0, \infty) \rightarrow \mathbb{R}$ be defined by $$ f_{1}(\mathrm{x})=\int_{0}^{\mathrm{x}} \prod_{\mathrm{j}=1}^{21}(\mathrm{t}-\mathrm{j})^{\mathrm{j}} \mathrm{dt}, \quad \mathrm{x}>0 $$ and $$ f_{2}(\mathrm{x})=98(\mathrm{x}-1)^{50}-600(\mathrm{x}-1)^{49}+2450, \mathrm{x}>0, $$ where, for any positive integer n and real numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}, \prod_{i=1}^{\mathrm{n}} \mathrm{a}_{i}$ denotes the product of $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$. Let $\mathrm{m}_{i}$ and $\mathrm{n}_{i}$, 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 $6 \mathrm{~m}_{2}+4 \mathrm{n}_{2}+8 \mathrm{~m}_{2} \mathrm{n}_{2}$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Application of Derivatives

Let

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

and

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

be defined by

f_{1} (x) = \int_{0}^{x} j = 1 \prod 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}, \dots, a_{n}, \prod_{i = 1}^{n} a_{i}

denotes the product of

a_{1}, a_{2}, \dots, 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 ____.

Answer: 6

f_{2}^{'} (x) = 98 \times 50 (x - 1)^{49} - 600 \times 49 (x - 1)^{48}

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

= 4900 (x - 1)^{48} (x - 7)

The derivative is zero at x = 1 and x = 7.

At x = 1:

(x - 1)^{48}

has even power, so the sign doesn't change. Not an extremum.

At x = 7: The factor

(x - 7)

changes sign from negative to positive, so x = 7 is a local minimum.

Therefore,

m_{2} = 1

and

n_{2} = 0

6 m_{2} + 4 n_{2} + 8 m_{2} n_{2} = 6 (1) + 4 (0) + 8 (1) (0) = 6

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer