CAT 2024 Slot 1QA Question 11

BasicsEasy

In the XY-plane, the area, in sq. units, of the region defined by the inequalities

y ≥ x + 4 and -4 ≤ x2 + y2 + 4(x - y) ≤ 0 is

Answer & solution

  • B

    π

  • C

  • D

Solution

Hard

Rewrite the quadratic inequality by completing the square — it becomes an annulus (region between two concentric circles). Then intersect that annulus with the half-plane yx+4y\ge x+4; the line happens to be tangent/chord-positioned so a clean fraction of the annulus survives.

1

Complete the square in the quadratic. Group xx and yy terms.

x2+y2+4x4y=(x+2)2+(y2)28 4(x+2)2+(y2)280(substitute) 4(x+2)2+(y2)28\begin{aligned} &x^2+y^2+4x-4y=(x+2)^2+(y-2)^2-8\\ &\Rightarrow\ -4\le (x+2)^2+(y-2)^2-8\le 0 \quad\text{(substitute)}\\ &\Rightarrow\ 4\le (x+2)^2+(y-2)^2\le 8 \end{aligned}
2

Interpret the region. An annulus centred at (2,2)(-2,2) with inner radius 22 and outer radius 222\sqrt2.

Centre (2,2),  rin=2,  rout=22\begin{aligned} &\text{Centre }(-2,2),\ \ r_{\text{in}}=2,\ \ r_{\text{out}}=2\sqrt2 \end{aligned}
3

Where does the line y=x+4y=x+4 sit? Distance from centre (2,2)(-2,2) to xy+4=0x-y+4=0.

d=(2)(2)+412+(1)2=02=0 the line passes through the centre(diameter line)\begin{aligned} &d=\dfrac{|(-2)-(2)+4|}{\sqrt{1^2+(-1)^2}}=\dfrac{0}{\sqrt2}=0\\ &\Rightarrow\ \text{the line passes through the centre} \quad\text{(diameter line)} \end{aligned}
4

Half the annulus survives. A line through the centre splits the annulus into two equal halves; yx+4y\ge x+4 keeps exactly one half.

Area=12(πrout2πrin2) Area=12π(84)(from step 2) Area=2π\begin{aligned} &\text{Area}=\tfrac12\left(\pi r_{\text{out}}^2-\pi r_{\text{in}}^2\right)\\ &\Rightarrow\ \text{Area}=\tfrac12\pi(8-4) \quad\text{(from step 2)}\\ &\Rightarrow\ \text{Area}=2\pi \end{aligned}
Area=2π sq. units\text{Area}=2\pi\ \text{sq. units}
CAT 2024 Slot 1 QA Q11: In the XY-plane, the area, in sq. units, of the region defined by the inequalities y ≥ x + 4 and -4 ≤ x — Solution | TheCATExam