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QJEEAdv19Maa9ec: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Maa9ec Solution: Let, , x > 0 Let x 1 < x 2 < x 3 < ... < x n < ... be all the…

Let, f(x)=sin⁡πxx2f(x) = {{\sin \pi x} \over {{x^2}}} , 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?

Multiple Choice Options

  • A ∣xn−yn∣ >1|{x_n} - {y_n}|\, > 1 for every n
  • B xn+1−xn >2{x_{n + 1}} - {x_n}\, > 2 for every n
  • C x1 < y1
  • D xn∈(2n, 2n+12){x_n} \in \left( {2n,\,2n + {1 \over 2}} \right) for every n

Step-by-Step Verified Solution

Given, f(x)=sin⁡(πx)x2f(x) = {{\sin (\pi x)} \over {{x^2}}} , x > 0

⇒f′(x)=x2πcos(πx)−2xsin⁡(πx)x4\Rightarrow f'(x) = {{{x^2}\pi cos(\pi x) - 2x\sin (\pi x)} \over {{x^4}}}

=2xcos⁡(πx)[xπ2−tan⁡(πx)]x4= {{2x\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^4}}}

=2cos⁡(πx)[xπ2−tan⁡(πx)]x3= {{2\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^3}}}

Since, for maxima and minima of f(x), f'(x) = 0

cos⁡(πx)=0\cos (\pi x) = 0 or tan⁡(πx)=πx2\tan (\pi x) = {{\pi x} \over 2} , (as x > 0)

∵\because cos⁡(πx)≠0⇒tan⁡(πx)=πx2\cos (\pi x) \ne 0 \Rightarrow \tan (\pi x) = {{\pi x} \over 2}

JEE Advanced 2019 Paper 2 Offline Mathematics - Application of Derivatives Question 13 English Explanation

∵\because f'(P_1^ - ) &lt; 0 and f'(P_1^ + ) &gt; 0

⇒\Rightarrow x=P1∈(1,32)x = {P_1} \in \left( {1,{3 \over 2}} \right)

is point of local minimum.

∵\because f'(P_2^ - ) &gt; 0 and f'(P_2^ + ) &lt; 0

⇒x=P2∈(2,52)\Rightarrow x = {P_2} \in \left( {2,{5 \over 2}} \right)

is point of local maximum.

From the graph, for points of maxima x1, x2, x3 .... it is clear that

{5 \over 2} - {x_1} &gt; {9 \over 2} - {x_2} &gt; {{13} \over 2} - {x_3} &gt; {{17} \over 2} - {x_4} .......

\Rightarrow {x_{n + 1}} - {x_n} &gt; 2 , ∀ n\forall \,n .

From the graph for points of minima y1, y2, y3 ....., it is clear that

{3 \over 2} - {y_1} &gt; {5 \over 2} - {x_1} &gt; {7 \over 2} - {y_2} &gt; {9 \over 2} - {x_2} ........

|{x_n} - {y_n}|\, &gt; 1 , ∀ n\forall \,n and x1 > (y1 + 1)

And x1∈(2, 52){x_1} \in \left( {2,\,{5 \over 2}} \right) , x2∈(4, 92){x_2} \in \left( {4,\,{9 \over 2}} \right) , x3∈(6, 132){x_3} \in \left( {6,\,{{13} \over 2}} \right) ..........

⇒\Rightarrow xn∈(2n,2n+ 12){x_n} \in \left( {2n,2n + \,{1 \over 2}} \right) , ∀ n\forall \,n .

Hence, options (a), (b) and (d) are correct.

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