Manipal Entrance Test 2015 Math Paper

We have,
AD = a + a cos 2θ and BC = 2BD = 2a sin 2θ
∴ Area of ΔABC is given by
Δ=

1

2

BC×AD =

2

2

a2(sin‌2‌θ+sin‌2‌θ‌cos‌2‌θ)
⇒ Δ=a2‌sin‌2‌θ+

1

2

a2‌sin‌4‌θ
On differentiating both sides w.r.t. θ, we get
dΔdθ=2a2‌cos‌2‌θ+2a2‌cos‌4‌θ
For maximum or minimum, put

dΔ

dθ

=0.
⇒ cos‌2‌θ=−cos‌4‌θ
⇒ 2θ=π−4θ
⇒ θ=

π

6

Now,

d2Δ

dθ2

=−2a2‌sin‌2‌θ×2 −2a2‌sin‌4‌θ×4
=−4a2‌sin‌2‌θ−8a2‌sin‌4‌θ
At θ=

π

6

,

d2Δ

dθ2

=−4a2‌sin‌

π

3

−8a2‌sin‌

2π

3

=−4a2×

√3

2

−8a2×

√3

2

=−6a2√3<0
Hence, Δ is maximum for θ=

π

6

.