JEE Advanced 2019 · Paper 2 · Mathematics Q5

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MathematicsMultiple correct+4 / -1Question 41 of 54

Let

$f(x) = \frac{\sin\pi x}{x^2},\quad x > 0.$

Let $x_1 < x_2 < x_3 < \cdots < x_n < \cdots$ be all the points of local maximum of $f$ and $y_1 < y_2 < y_3 < \cdots < y_n < \cdots$ be all the points of local minimum of $f$ . Then which of the following options is/are correct?

x_1 < y_1

x_{n+1} - x_n > 2

for every

n

x_n\in\left(2n, 2n + \frac12\right)

for every

n

|x_n - y_n| > 1

for every

n

Answer and worked solution

Official answer: B, C, D

Critical points.
$f'(x)=\frac{\pi x\cos\pi x-2\sin\pi x}{x^3}=0\iff\tan\pi x=\frac{\pi x}2.$

Behaviour. $f$ falls from $+\infty$ on $(0,1)$ . It is negative on each interval $(2n-1,2n)$ , with its minimum $y_n$ in $(2n-1,\,2n-\frac12)$ . It is positive on $(2n,2n+1)$ , with its maximum $x_n$ in $(2n,\,2n+\frac12)$ .

Where the roots sit. In each branch the root of $\tan\pi x=\frac{\pi x}2$ moves closer to the asymptote at $m+\frac12$ as $m$ grows. Write $x_n=2n+\delta_n$ and $y_n=2n-1+\epsilon_n$ . Then $\delta_n$ increases with $n$ , and $\epsilon_n<\delta_n$ .

The options.

(A) $x_1\in(2,2.5)$ while $y_1\in(1,1.5)$ , so $x_1>y_1$ . False.
(B) $x_{n+1}-x_n=2+\delta_{n+1}-\delta_n>2$ . True.
(C) $x_n\in\left(2n,2n+\frac12\right)$ . True.
(D) $x_n-y_n=1+\delta_n-\epsilon_n>1$ . True.

(B), (C), (D)

Topics:Maxima and minima Monotonicity

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JEE Advanced 2019 · Paper 2, Mathematics question 5