In the X – Y plane, the area of the region bounded by the graph of | x + y | + | x – y | = 4 is(2005)

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Solution

The correct option is C 16
Let x $\geq$ 0, y $\geq$ 0 and x $\geq$ y

Then, | x + y | + | x – y | = 4

$\to$ x + y + x – y = 4 $\to$ x = 2

And in case x $\geq$ 0, y $\geq$ 0, x $\leq$ y

x + y + y – x = 4 $\to$ y = 2

Area in the first quadrant is 4.

By symmetry, total area 4 $\times$ 4 = 16 unit.

Suggest Corrections

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x = \pm 2, y \pm 2

| 5 s i n x |

x = 4 π

y = | x + 4 |

y = | x + 4 |

x = - 6

x = 0

x \geq 0

y \geq 0

[x] + [y] = 4

List-I	List-II
1. Area of region bounded by $y = 2 x - x^{2}$ and $x -$ axis	a. $\frac{1}{3}$
2. Area of the region ${(x, y) : x^{2} \leq y \leq \| x \|}$	b. $\frac{1}{2}$
3. Area bounded by $y = x$ and $y = x^{3}$	c. $\frac{2}{3}$
4. Area bounded by $y = x \| x \|$ , $x$ -axis and $x = - 1, x = 1$	d. $\frac{4}{3}$

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