JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 3)
Let, $$f(x) = {{\sin \pi x} \over {{x^2}}}$$, 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?
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?
$$|{x_n} - {y_n}|\, > 1$$ for every n
$${x_{n + 1}} - {x_n}\, > 2$$ for every n
x1 < y1
$${x_n} \in \left( {2n,\,2n + {1 \over 2}} \right)$$ for every n
Explanation
Given, $$f(x) = {{\sin (\pi x)} \over {{x^2}}}$$, x > 0
$$ \Rightarrow f'(x) = {{{x^2}\pi cos(\pi x) - 2x\sin (\pi x)} \over {{x^4}}}$$
$$ = {{2x\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^4}}}$$
$$ = {{2\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^3}}}$$
Since, for maxima and minima of f(x), f'(x) = 0
$$\cos (\pi x) = 0$$ or $$\tan (\pi x) = {{\pi x} \over 2}$$, (as x > 0)
$$ \because $$ $$\cos (\pi x) \ne 0 \Rightarrow \tan (\pi x) = {{\pi x} \over 2}$$
$$ \because $$ $$f'(P_1^ - ) < 0$$ and $$f'(P_1^ + ) > 0$$
$$ \Rightarrow $$ $$x = {P_1} \in \left( {1,{3 \over 2}} \right)$$
is point of local minimum.
$$ \because $$ $$f'(P_2^ - ) > 0$$ and $$f'(P_2^ + ) < 0$$
$$ \Rightarrow x = {P_2} \in \left( {2,{5 \over 2}} \right)$$
is point of local maximum.
From the graph, for points of maxima x1, x2, x3 .... it is clear that
$${5 \over 2} - {x_1} > {9 \over 2} - {x_2} > {{13} \over 2} - {x_3} > {{17} \over 2} - {x_4}$$ .......
$$ \Rightarrow {x_{n + 1}} - {x_n} > 2$$, $$\forall \,n$$.
From the graph for points of minima y1, y2, y3 ....., it is clear that
$${3 \over 2} - {y_1} > {5 \over 2} - {x_1} > {7 \over 2} - {y_2} > {9 \over 2} - {x_2}$$ ........
$$|{x_n} - {y_n}|\, > 1$$, $$\forall \,n$$ and x1 > (y1 + 1)
And $${x_1} \in \left( {2,\,{5 \over 2}} \right)$$, $${x_2} \in \left( {4,\,{9 \over 2}} \right)$$, $${x_3} \in \left( {6,\,{{13} \over 2}} \right)$$ ..........
$$ \Rightarrow $$ $${x_n} \in \left( {2n,2n + \,{1 \over 2}} \right)$$, $$\forall \,n$$.
Hence, options (a), (b) and (d) are correct.
$$ \Rightarrow f'(x) = {{{x^2}\pi cos(\pi x) - 2x\sin (\pi x)} \over {{x^4}}}$$
$$ = {{2x\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^4}}}$$
$$ = {{2\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^3}}}$$
Since, for maxima and minima of f(x), f'(x) = 0
$$\cos (\pi x) = 0$$ or $$\tan (\pi x) = {{\pi x} \over 2}$$, (as x > 0)
$$ \because $$ $$\cos (\pi x) \ne 0 \Rightarrow \tan (\pi x) = {{\pi x} \over 2}$$
$$ \because $$ $$f'(P_1^ - ) < 0$$ and $$f'(P_1^ + ) > 0$$
$$ \Rightarrow $$ $$x = {P_1} \in \left( {1,{3 \over 2}} \right)$$
is point of local minimum.
$$ \because $$ $$f'(P_2^ - ) > 0$$ and $$f'(P_2^ + ) < 0$$
$$ \Rightarrow x = {P_2} \in \left( {2,{5 \over 2}} \right)$$
is point of local maximum.
From the graph, for points of maxima x1, x2, x3 .... it is clear that
$${5 \over 2} - {x_1} > {9 \over 2} - {x_2} > {{13} \over 2} - {x_3} > {{17} \over 2} - {x_4}$$ .......
$$ \Rightarrow {x_{n + 1}} - {x_n} > 2$$, $$\forall \,n$$.
From the graph for points of minima y1, y2, y3 ....., it is clear that
$${3 \over 2} - {y_1} > {5 \over 2} - {x_1} > {7 \over 2} - {y_2} > {9 \over 2} - {x_2}$$ ........
$$|{x_n} - {y_n}|\, > 1$$, $$\forall \,n$$ and x1 > (y1 + 1)
And $${x_1} \in \left( {2,\,{5 \over 2}} \right)$$, $${x_2} \in \left( {4,\,{9 \over 2}} \right)$$, $${x_3} \in \left( {6,\,{{13} \over 2}} \right)$$ ..........
$$ \Rightarrow $$ $${x_n} \in \left( {2n,2n + \,{1 \over 2}} \right)$$, $$\forall \,n$$.
Hence, options (a), (b) and (d) are correct.
Comments (0)
