KVPY SX SB Exam 2018 Question Paper

Question : 9 of 120

Marks: +1, -0

A rectangle with its sides parallel to the x-axis and y-axis is inscribed in the region bounded by the curves

y = x^2 - 4

and

2y = 4 - x^2

. The maximum possible area of such a rectangle is closest to the integer

Solution:

\A=2x[(2-x^2/2)-(x^2-4)]

=x(12-3x^2)=12x-3x^2

{dA}/{dx}=0⇒12-9x^2=0

⇒x^2=4/3⇒x=2/√3

\A=3×2/√3(4-4/3) =48/{3√3}

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