UPSC NDA I 2014 Mathematics Paper

lim

x→ π 2

f(x)=

1−sinx

(π−2x)2

(

0

0

‌form)
⇒

lim

x→ π 2

f(x)=

−cosx

−4(π−2x)

⇒

lim

x→ π 2

f(x)=

sinx

8

=

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
Areamax=2ab
We know that area of ellipse = abπ
Desired area =2ab−abπ=ab(2−π)