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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 by Veteran (431k points)
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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.
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