# ISI2014-DCG-28

1 vote
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The area enclosed by the curve $\mid\: x \mid + \mid y \mid =1$ is

1. $1$
2. $2$
3. $\sqrt{2}$
4. $4$
in Calculus
recategorized

Answer: $B.$ $2$

$x+y=1$. when$x>0$ and$y>0$

$-x+y=1$ when $x<0$ and $y>40$

$-x-y=1$ when both $x$ and $y$ are $<0$

$x-y=1$ when $x>0$ and $y<0$

Now, on plotting these equations,

The obtained coordinates would be:

$(1,0) (0,1) (-1,0) (0,-1)$

On joining these co-ordinates, a square of side $\sqrt2$ units would be obtained.

Therefore, area enclosed $=$ Area of square $=(\sqrt2)^2 = 2$ units

Therefore, option $B$ is the right answer.

Sorry for the bad handwriting!

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