GATE Overflow Test Series | Mock GATE | Test 1 | Question: 10

Question

The area of the closed region by the equation $\mid x \mid + \mid y \mid = 4$ in the two - dimensional plane is?

• A. $32\:\pi$
• B. $16$
• C. $8$
• D. $32$

Answer 1

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

Given that: $\mid x \mid  + \mid y \mid = 4$

There are four case, for which give equation is satisfied.

• $x + y = 4 \quad \rightarrow (1)$
• $-x + y = 4\quad \rightarrow (2)$
• $x - y = 4\quad \rightarrow (3)$
• $-x - y = 4\quad \rightarrow (4)$

Area enclosed by four lines $ = \triangle\: AOB + \triangle \:BOC + \triangle\: AOD + \triangle\: COD $

We know that area of $\triangle = \dfrac{1}{2} \times \text{base} \times \text{height}$

Total area $ = \left(\dfrac{1}{2}\times 4 \times 4\right) \times 4 = 32.$

So, the correct answer is $(D).$

Comments

@gatecse @Sachin Mittal 1 
I used following approach, please tell what is wrong with it.

|x| + |y| = 4

|y| = 4 - |x|

If x >= 0 : |y| = 4 - x

Y = 4 - x              if 4 - x >= 0 => x <= 4   and x >= 0(base condition)      — (1)

Y = -4 + x            if 4 - x < 0 => x > 4          —(2)

If x < 0 : |y| = 4 + x 

Y = 4 + x             if 4+x >= 0 => x >= -4     and x <0 (base condition)    —(3)

Y = -4-x              if 4 + x < 0 => x < -4      —-(4)

When we plot abv 4 equatioins in their ranges, we get a graph like below : 

Please let me know what is wrong with this approach.

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$4-x$ is same as $|y|$. They both are EQUAL.

$|y|$ is ALWAYS $\geq0$ hence $4-x$ is always $\geq0$.

Y = -4 + x            if 4 - x < 0 => x > 4          —(2)

This is wrong in your method. 

If you really want to make cases, make two cases for $y$, not for $4-x$.