Answer

Verified

482.4k+ views

Hint – In this question we have been given two modulus functions, diagrammatic representation of the two given functions can help us in understanding the basic outline sketch of the problem that we are dealing with. Use the basic definitions of $\left| a \right| \leqslant x$ to understand the region which we are dealing with.

Complete step-by-step answer:

Given equation is

$\left| {x - y} \right| \leqslant 2$ and $\left| {x + y} \right| \leqslant 2$

Now according to property of modulus if$\left| a \right| \leqslant x$, then $a \leqslant x$ and $ - a \leqslant x$.

Then convert the equations according to this property we have,

$\left| {x - y} \right| \leqslant 2$

$x - y \leqslant 2$………………… (1)

And $ - \left( {x - y} \right) \leqslant 2$

$ \Rightarrow - x + y \leqslant 2$………………….. (2)

Now, for equation $\left| {x + y} \right| \leqslant 2$

$x + y \leqslant 2$ …………………………. (3)

And $ - \left( {x + y} \right) \leqslant 2$

$ \Rightarrow - x - y \leqslant 2$ ……………………. (4)

Now plot all these equations

Equation (1) represents line CD (see figure) as x coefficient is positive and y coefficient is negative therefore line lie in fourth quadrant and the intersection point of this line with x and y axis are (2, 0) and (0, -2) (see figure).

Equation (2) represents line AB (see figure) as x coefficient is negative and y coefficient is positive therefore line lie in second quadrant and the intersection point of this line with x and y axis are (-2, 0) and (0, 2) (see figure).

Equation (3) represents line DA (see figure) as x coefficient is positive and y coefficient is positive therefore line lie in the first quadrant and the intersection point of this line with x and y axis are (2, 0) and (0, 2) (see figure).

Equation (4) represents line BC (see figure) as x coefficient is negative and y coefficient is negative therefore line lie in third quadrant and the intersection point of this line with x and y axis are (-2, 0) and (0, -2) (see figure).

Now all the lines are less than or equal to 2 so the region bounded by the curves is represented in the figure.

Now calculate the distance AB.

The distance between two points (a, b) and (c, d) is given as $d = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $

Let the point A = (a, b) = (0, 2), point B = (c, d) = (-2, 0), point C = (e, f) = (0, -2), point D = (g, h) = (2, 0)

Now the distance $AB = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} = \sqrt {{{\left( { - 2 - 0} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $BC = \sqrt {{{\left( {e - c} \right)}^2} + {{\left( {f - d} \right)}^2}} = \sqrt {{{\left( {0 + 2} \right)}^2} + {{\left( { - 2 - 0} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $CD = \sqrt {{{\left( {g - e} \right)}^2} + {{\left( {h - f} \right)}^2}} = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {0 + 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $DA = \sqrt {{{\left( {g - a} \right)}^2} + {{\left( {h - b} \right)}^2}} = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now as we see all the distances are equal

$ \Rightarrow AB = BC = CD = DA$

So this is the condition of the square.

So, the region represented by given lines is a square with side length $2\sqrt 2 $ sq. unit.

Hence option (A) is correct.

Note – Whenever we face such types of problems the key concept is to have the basic understanding of modulus function, modulus function involving inequalities, definitions of rhombus, square and the basic area formula. This along with the curve outline sketch will help you reach the right answer.

Complete step-by-step answer:

Given equation is

$\left| {x - y} \right| \leqslant 2$ and $\left| {x + y} \right| \leqslant 2$

Now according to property of modulus if$\left| a \right| \leqslant x$, then $a \leqslant x$ and $ - a \leqslant x$.

Then convert the equations according to this property we have,

$\left| {x - y} \right| \leqslant 2$

$x - y \leqslant 2$………………… (1)

And $ - \left( {x - y} \right) \leqslant 2$

$ \Rightarrow - x + y \leqslant 2$………………….. (2)

Now, for equation $\left| {x + y} \right| \leqslant 2$

$x + y \leqslant 2$ …………………………. (3)

And $ - \left( {x + y} \right) \leqslant 2$

$ \Rightarrow - x - y \leqslant 2$ ……………………. (4)

Now plot all these equations

Equation (1) represents line CD (see figure) as x coefficient is positive and y coefficient is negative therefore line lie in fourth quadrant and the intersection point of this line with x and y axis are (2, 0) and (0, -2) (see figure).

Equation (2) represents line AB (see figure) as x coefficient is negative and y coefficient is positive therefore line lie in second quadrant and the intersection point of this line with x and y axis are (-2, 0) and (0, 2) (see figure).

Equation (3) represents line DA (see figure) as x coefficient is positive and y coefficient is positive therefore line lie in the first quadrant and the intersection point of this line with x and y axis are (2, 0) and (0, 2) (see figure).

Equation (4) represents line BC (see figure) as x coefficient is negative and y coefficient is negative therefore line lie in third quadrant and the intersection point of this line with x and y axis are (-2, 0) and (0, -2) (see figure).

Now all the lines are less than or equal to 2 so the region bounded by the curves is represented in the figure.

Now calculate the distance AB.

The distance between two points (a, b) and (c, d) is given as $d = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $

Let the point A = (a, b) = (0, 2), point B = (c, d) = (-2, 0), point C = (e, f) = (0, -2), point D = (g, h) = (2, 0)

Now the distance $AB = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} = \sqrt {{{\left( { - 2 - 0} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $BC = \sqrt {{{\left( {e - c} \right)}^2} + {{\left( {f - d} \right)}^2}} = \sqrt {{{\left( {0 + 2} \right)}^2} + {{\left( { - 2 - 0} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $CD = \sqrt {{{\left( {g - e} \right)}^2} + {{\left( {h - f} \right)}^2}} = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {0 + 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now the distance $DA = \sqrt {{{\left( {g - a} \right)}^2} + {{\left( {h - b} \right)}^2}} = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 $

Now as we see all the distances are equal

$ \Rightarrow AB = BC = CD = DA$

So this is the condition of the square.

So, the region represented by given lines is a square with side length $2\sqrt 2 $ sq. unit.

Hence option (A) is correct.

Note – Whenever we face such types of problems the key concept is to have the basic understanding of modulus function, modulus function involving inequalities, definitions of rhombus, square and the basic area formula. This along with the curve outline sketch will help you reach the right answer.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Who was the Governor general of India at the time of class 11 social science CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference Between Plant Cell and Animal Cell