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 $2 \mathrm{~m}_{1}+3 \mathrm{n}_{1}+\mathrm{m}_{1} \mathrm{n}_{1}$ 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

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

is ____.

Answer: 57

Solution

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

The derivative changes sign at x = 1, 2, 3, ..., 21.

At odd powers (x = 1, 3, 5, ..., 21), the sign changes. At even powers (x = 2, 4, 6, ..., 20), the sign does not change (touches but doesn't cross).

Local minima occur at x = 1, 5, 9, 13, 17, 21 (where derivative goes from negative to positive). These are at positions

4 m + 1

for m = 0, 1, 2, 3, 4, 5. So

m_{1} = 6

.

Local maxima occur at x = 3, 7, 11, 15, 19 (where derivative goes from positive to negative). These are at positions

4 m - 1

for m = 1, 2, 3, 4, 5. So

n_{1} = 5

.

Therefore,

2 m_{1} + 3 n_{1} + m_{1} n_{1} = 2 (6) + 3 (5) + 6 (5) = 12 + 15 + 30 = 57

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

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

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