Question

The area of the smaller region in which the curve

y =where [.] denotes the greatest integer function, divides the circle (x - 2)² + (y + 1)² = 4, is equal to

• A. $\frac { 2 \pi - 3 \sqrt { 3 } } { 3 }$sq. units
• B. $\frac { 3 \sqrt { 3 } - \pi } { 3 }$ sq. units
• C. $\frac { 4 \pi - 3 \sqrt { 3 } } { 3 }$sq. units
• D. $\frac { 5 \pi - 3 \sqrt { 3 } } { 3 }$ sq. units

Answer 1

Answer: (C)

$\frac { 4 \pi - 3 \sqrt { 3 } } { 3 }$sq. units

Answer 2

Circle has (2, -1) as it's center and radius of this circle is 2. Thus, if P(x, y) be any point on it then x e [0, 4].

Let g(x) =$\frac { x ^ { 3 } } { 100 } + \frac { x } { 50 }$

⇒ g'(x)= $\frac { 3 x ^ { 3 } } { 100 } + \frac { 1 } { 50 }$ > 0 ∀ x ∈ (0, 4]

Thus g(x) is increasing in [0, 4]. g(0) = 0, g(4) = $\frac { 18 } { 25 }$ .

Hence g(x) ∈ $\left[ 0 , \frac { 18 } { 25 } \right]$ ∀ x ∈ [0, 4]

⇒ [g(x)] = 0 ∀ x ∈ [0, 4]

Thus y = $\left[ \frac { x ^ { 3 } } { 100 } + \frac { x } { 50 } \right]$, simply represents the x-axis.

CA = CB = 2, CD = 1

⇒ cos θ = ⇒ ⇒ ∠ACB =

$\Delta _ { \mathrm { ACB } } = \frac { 1 } { 2 } \cdot 2 ^ { 2 } \cdot \sin \frac { 2 \pi } { 3 } = \sqrt { 3 }$ sq. units.

Area of sector ACB = $\frac { 1 } { 2 } \cdot 2 ^ { 2 } \cdot \frac { 2 \pi } { 3 } = \frac { 4 \pi } { 3 }$ sq. units

Thus area of smaller segment

$= \left( \frac { 4 \pi } { 3 } - \sqrt { 3 } \right) = \left( \frac { 4 \pi - 3 \sqrt { 3 } } { 3 } \right)$ sq. units.