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NDA Math Area Under the Curves

DIRECTIONS: (For the next two (02) items that follow):

Consider an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

[NDA I 2014]

Question : 27 of 27

Marks: +1, -0

Solution:

\underset{x\to \pi/2}{\lim} f(x) = \frac{1-\sin x}{(\pi-2x)^2} \text{(0/0form)}

\Rightarrow \underset{x\to \pi/2}{\lim} f(x) = \frac{-\cos x}{-4(\pi-2x)}

\Rightarrow \underset{x\to \pi/2}{\lim} f(x) = \frac{\sin x}{8} = \frac{1}{8}

Let ∠AOM = θ, then point A be (acosθ,sinθ)

The length of rectangle = 2acosθ

The width of rectangle = 2bsinθ

Area = 4absinθcosθ = 2absin2θ

For maximum sin2θ = 1

\text{Area}_{\max} = 2ab

We know that area of ellipse = abπ

Desired area

= 2ab - ab\pi = ab (2 - \pi)

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