CAT 2020 Set-1 | Question: 61

Question

The area of the region satisfying the inequilities $\left | x \right |-y\leq 1,y\geq 0$ and $y\leq 1$ is

Answer 1

Given that, $|x| – y \leq 1 ; y \geq 0$ and $y \leq 1.$

We know that, $|x| =  \left\{\begin{matrix} x ; & x \geq 0 & \\ -x ; & x < 0 & \end{matrix}\right.$

Let us assume $|x| – y = 1 ; y = 0 ; y = 1$

$ \Rightarrow y = |x| – 1 \quad \longrightarrow (1)$

We know that, $y = |x|$ graph.
 

Then, $y = |x| – 1$ graph will be
 

So, final graph for equation $(1).$

We know that, $\text{area of a trapezium} = \frac{1}{2} \text{(sum of parallel sides)} \times \text{Distance between them}$

So, the required area $ = \frac{1}{2} (4+2) \times 1 = \frac{6}{2} = 3 \; \text{unit square.}$
$$\text{(OR)}$$
$\text{The area of ABCD = Area of PQCD – Area of PAD – Area of QBC}$

$\qquad = (4 \times 1) – \frac{1}{2} \times 1 \times 1 – \frac{1}{2} \times 1 \times 1$

$\qquad = 4 – \frac{1}{2} – \frac{1}{2} $

$\qquad = 4 – \left( \frac{2}{2} \right)$

$\qquad = 4 – 1 $

$\qquad = 3 \; \text{unit square}.$

Correct Answer$: 3$