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Question

# Let f(x)=sinπxx2, 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?

A
x1<y1
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B
xn+1xn>2 for every n
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C
xn(2n,2n+12) for every n
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D
|xnyn|>1 for every n
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Solution

## The correct option is D |xn−yn|>1 for every nf(x)=sinπxx2 f′(x)=πx2cosπx−2xsinπxx4=2cosπx(πx2−tanπx)x3 where x≠0 For maxima and minima: f′(x)=0 ⇒cosπx(πx2−tanπx)=0 Let g(x)=cosπx or, h(x)=tanπx−πx2 From the above graph we can observe that product of both g(x) and h(x) changes its sign at given interval ∴πx=(2n+1)π2 ⇒x=2n+12, n∈I From the graph, we can see that for all x=2n+12, doesn't change sign. Also we can see x1 is point of maxima and y1 is point of minima and x1>y1 Same is the case for all x satisfying πx2=tanπx where yn∈(2n−1,2n−12) ∀ n=1,2,3.... and xn∈(2n,2n+12) ∀ n=1,2,3.... ∴|xn−yn|>1 for every n x1>y1 xn+1−yn+1>1 and yn+1−xn>1 ⇒xn+1−xn>2

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