Let $\mathrm{f}(\mathrm{x})=\frac{\sin \pi \mathrm{x}}{\mathrm{x}^{2}}, \mathrm{x}>0$ Let $\mathrm{x}_{1}<\mathrm{x}_{2}<\mathrm{x}_{3}<\ldots<\mathrm{x}_{\mathrm{n}}<\ldots$ be all the points of local maximum of f and $\mathrm{y}_{1}<\mathrm{y}_{2}<\mathrm{y}_{3}<\ldots<\mathrm{y}_{\mathrm{n}}<\ldots$ be all the points of local minimum of f. Then which of the following options is/are correct ?

(A) $\left|\mathrm{x}_{\mathrm{n}}-\mathrm{y}_{\mathrm{n}}\right|>1$ for every n; (C) $\mathrm{x}_{\mathrm{n}} \in\left(2 \mathrm{n}, 2 \mathrm{n}+\frac{1}{2}\right)$ for every n; (D) $\mathrm{x}_{\mathrm{n}+1}-\mathrm{x}_{\mathrm{n}}>2$ for every n

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Application of Derivatives

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Let

f (x) = \frac{s i n π x}{x ^{2}}, x > 0

Let

x_{1} < x_{2} < x_{3} < \dots < x_{n} < \dots

be all the points of local maximum of f and

y_{1} < y_{2} < y_{3} < \dots < y_{n} < \dots

be all the points of local minimum of f. Then which of the following options is/are correct ?

(A) $∣ x_{n} - y_{n} ∣ > 1$ for every n✓ Correct
(B) $x_{1} < y_{1}$
(C) $x_{n} \in (2 n, 2 n + \frac{1}{2})$ for every n✓ Correct
(D) $x_{n + 1} - x_{n} > 2$ for every n✓ Correct

Answer: (A), (C), (D)

Solution

We have

f^{'} (x) = \frac{2 x c o s π x ( \frac{π x}{2} - t a n π x )}{x ^{4}}

. The critical points occur when

cos π x = 0

(giving

x = (2 k + 1) /2

) or when

\frac{π x}{2} = tan π x

. Analyzing the graph of

y = \frac{π x}{2}

and

y = tan π x

, we see that in each interval

(2 n, 2 n + 1)

, there is one intersection point

x_{n} \in (2 n, 2 n + 1/2)

(local max) and one at

y_{n} \in (2 n + 1/2, 2 n + 1)

(local min). From the graph:

∣ x_{n} - y_{n} ∣ > 1

x_{1} > y_{1}

x_{n} \in (2 n, 2 n + 1/2)

, and

x_{n + 1} - x_{n} > 2

. Options (A), (C), and (D) are correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Application of Derivatives

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