Let fx=sin⁡πxx2, x>0. Let x1<x2<x3<…<xn<… be all the points of local maximum of f and y1<y2<y3<…<yn<… be all the points of local minimum of f. Then which of the following options is/are correct?

JEE Advanced · Mathematics · 25. AOD

Let $f (x) = \frac{\sin π x}{x^{2}}, x > 0 .$
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 $n$
B $x_{1} < y_{1}$
C $x_{n} \in (2 n, 2 n + \frac{1}{2})$ for every $n$
D $x_{n + 1} - x_{n} > 2$ for every $n$

Verified Solution

Answer & Solution

Correct Answer

(D) $x_{n + 1} - x_{n} > 2$ for every $n$

Step-by-step Solution

Detailed explanation

f' (x) = \frac{2 x \cos π x (\frac{π x}{2} - \tan π x)}{x^{4}}

…(i)

\Rightarrow

for maxima/minima

f^{'} (x) = 0

\begin{matrix} \Rightarrow \cos π x = 0 & o r \frac{π x}{2} = \tan π x \end{matrix}

\begin{matrix} ∵ \cos π x \neq 0 & \{∵ \tan π x will not be defined\} \end{matrix}

∴

maxima/minima will occur
Where

\tan (π x) = \frac{π x}{2}

Case

I :

(i)

For example
In

x \in (2 n, 2 n + \frac{1}{2})

at point

P_{2}, x \in (2, \frac{5}{2})

\cos (π x) > 0

and at

P_{2}^{+}, \tan (π x) > \frac{π x}{2}

P_{2}^{-}, \tan (π x) < \frac{π x}{2}

hence from equation (i)
in

x \in (2 n, 2 n + \frac{1}{2}), f^{'} (x)

goes from positive to negative
hence

P_{2}

is maxima
Similarly

P_{2}, P_{4}, P_{6}

… are point of maxima and all lies in

x \in (2 n, 2 n + \frac{1}{2})

Case

I I

x \in (2 n + 1, 2 n + \frac{3}{2})

(i i)

for example at

P_{1}, x \in (1, \frac{3}{2})

\cos (π x) < 0

P_{1}^{+}, \tan (π x) > \frac{π x}{2}

and at

P_{1}^{-}, \tan (π x) < \frac{π x}{2}

hence from equation (i)
in

x \in (2 n + 1, 2 n + \frac{3}{2}), f^{'} (x)

goes from negative to positive

\Rightarrow

so, for the minima

y_{1} : 1 < y_{1} < \frac{3}{2}

y_{2} : 3 < y_{2} < \frac{7}{2}

y_{3} : 5 < y_{3} < \frac{11}{2}

and for the maxima

x_{1} : 2 < x_{1} < \frac{5}{2}

x_{2} : 4 < x_{2} < \frac{9}{2}

x_{3} : 6 < x_{3} < \frac{13}{2}

Hence

x_{n} > y_{n}

…

(a)

\Rightarrow

now,

x_{1} > y_{1}

\tan (π x_{1}) > \tan (π y_{1})

\tan (π x_{1}) > \tan (π + π y_{1})

\tan (π x_{1}) > \tan (π (1 + y_{1}))

π x_{1} > π (1 + y_{1})

x_{1} > 1 + y_{1}

Similarly,

x_{n} > 1 + y_{n}

x_{n} - y_{n} > 1

…(b)

|x_{n} - y_{n}| > 1

\Rightarrow y_{2} > x_{1}

\tan (π y_{2}) > \tan (π x_{1})

\tan (π y_{2}) > \tan (π + π x_{1}) \Rightarrow y_{2} > x_{1} + 1

\Rightarrow y_{n + 1} > x_{n} + 1

Therefore

x_{n} < y_{n + 1} < x_{n + 1} \Rightarrow y_{n + 1} - x_{n} > 1

…(c)
Now consider

x_{n + 1} - x_{n} = (x_{n + 1} - y_{n + 1}) + (y_{n + 1} - x_{n})

From (b) and (c)

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

let

f_{1} : R \to R, f_{2} : [0, \infty) \to R, f_{3} : R \to R a n d f_{4} : R \to [0, \infty)

be defined by

f_{1} (x) = \{\begin{matrix} |x| & i f & x < 0 \\ e^{x} & i f & x \geq 0 \end{matrix}; f_{2} (x) = x^{2}; f_{3} (x) = \{\begin{matrix} \sin x & i f & x < 0 \\ x & i f & x \geq 0 \end{matrix} a n d f_{4} (x) = \{\begin{matrix} f_{2} (f_{1} (x)) & i f & x < 0 \\ f_{2} (f_{1} (x)) - 1 & i f & x \geq 0 \end{matrix}

	List – I		List – II
(A)	$f_{4}$ is	(P)	Onto but not one – one
(B)	$f_{3}$ is	(Q)	neither continuous nor one-one
(C)	$f_{2}$ of $f_{1}$ is	(R)	differentiable but not one-one
(D)	$f_{2}$ is	(S)	continuous and one-one

JEE Advanced 2014 Medium

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Let fx=sin⁡πxx2, x>0.Let x1<x2<x3<…<xn<… be all the points of local maximum of f and y1<y2<y3<…<yn<… be all the points of local minimum of f.Then which of the following options is/are correct?

Step-by-step Solution

Let $f (x) = \frac{\sin π x}{x^{2}}, x > 0 .$
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?