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Question

For the curve f(x)=11+x2, let two points on it be A(α,f(α)),B(1α,f(1α))(α>0). Find the minimum area bounded by the line segments OA, OB and f(x), where 'O' is the origin.

A
(π1)2
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B
π2
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C
(π2)2
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D
Maximum area is always infinite
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Solution

The correct option is A (π1)2

The shaded area in the graph is the required area that has to be maximised
Point A (α,11+α2)
Point B (1α,α21+α2)
Thus the required area to be maximised is,
A=α1/αf(x)dx(Area of triangle under segment OB) (Area of triangle under segment OA)
f(x)dx=tan1x
A=tan1xα1/α(12×1α×α21+α2)(12×α×11+α2)
A=tan1α+tan11αα1+α2
tan1α+tan11α=π2
A=π2α1+α2
This is maximum when α1+α2 is minimum.
ddα(α1+α2)=0
1+α2α(2α)1+α2=0
α=±1
Maximum value occurs at α=1
Therefore minimum value of bounded area =π212

358017_132369_ans_eddd2d35f51546d0ad5647ae5919993c.png

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