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#### Question 1:

Mark on a number line the positions of the numbers satisfying each of the conditions below:

(i) 1 < x < 3

(ii) −3 < x < −1

(iii) −3 < x < 1

(iv) −1 < x < 3

(i)

1 < x < 3

The positions of the numbers satisfying the above condition can be shown on the number line as follows: (ii)

3 < x < 1

The positions of the numbers satisfying the above condition can be shown on the number line as follows: (iii)

3 < x < 1

The positions of the numbers satisfying the above condition can be shown on the number line as follows: (iv)

1 < x < 3

The positions of the numbers satisfying the above condition can be shown on the number line as follows: #### Question 1:

For each of the pictures below, give the algebraic condition satisfied by all numbers in the portion marked: (1)

The algebraic condition satisfied by all the numbers in the marked portion is 1 < x < 2.

(2)

The algebraic condition satisfied by all the numbers in the marked portion is 4 < x < 1.

(3)

The algebraic condition satisfied by all the numbers in the marked portion is 2 < x < 3.

#### Question 1:

Mark each pair of numbers given below on the number line and check whether the distance between them is equal to the number got by subtracting the smaller from the larger:

(i) 1, −5

(ii) −3, −4

(iii) −1, 1

(iv) 0, 4

(v) 0, −4

(vi) (i)

1, 5 can be represented on the number line as follows: Distance between 5 and 1 = 6 units

Required difference = 1 − (5) = 1 + 5 = 6

Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

(ii)

3, 4 can be represented on the number line as follows: Distance between 4 and 3 = 1 unit

Required difference = 3 − (4) = 3 + 4 = 1

Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

(iii)

1, 1 can be represented on the number line as follows: Distance between 1 and 1 = 2 units

Required difference = 1 − (1) = 1 + 1 = 2

Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

(iv)

0, 4 can be represented on the number line as follows: Distance between 0 and 4 = 4 units

Required difference = 4 − 0 = 4

Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

(v)

0, 4 can be represented on the number line as follows: Distance between 4 and 0 = 4 units

Required difference = 0 − (4) = 0 + 4 = 4

Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

(vi) can be represented on number line as follows: Distance between Required difference = Yes, the distance between the given numbers is equal to the number got by subtracting the smaller number from the larger number.

#### Question 1:

Find x satisfying each of the following equations:

(i) (ii) (iii) (iv) (v) (vi) (i) If x > 3 then = x − 3

x − 3 = 2

x = 2 + 3 = 5

If x < 3 then = (x − 3) = 3 − x

3 − x = 2

x = 3 2 = 1

x = 1 and 5

(ii) If x > 2 then = x + 2

x + 2 = 1

x = 1 2 = 1

If x < 2 then = (x + 2) = x 2

⇒ −x 2 = 1

⇒ −x = 1 + 2

x = 3

x = 1 and 3

(iii) If x < 1 then (x − 1) = (x − 3)

x − 1 = x − 3

This is an invalid equation.

If 1 < x < 3 then (x − 1) = (x − 3)

x − 1 = x + 3

x + x = 3 + 1

2x = 4

x = 2

If x > 3 then (x − 1) = (x − 3)

This is again an invalid equation.

Thus, x = 2 satisfies the given equation.

(iv) If x < 3 then (x − 3) = (x − 4)

x − 3 = x − 4

This is an invalid equation.

If 3 < x < 4 then (x − 3) = (x − 4)

x − 3 = x + 4

x + x = 4 + 3

2x = 7

x = If x > 4 then (x − 3) = (x − 4)

This is again an invalid equation.

Thus, x = satisfies the given equation.

(v) If x < 2 then (x + 2) = (x − 5)

x + 2 = x − 5

This is an invalid equation.

If 2 < x < 5 then (x + 2) = (x − 5)

x + 2 = x + 5

x + x = 5 2

2x = 3

x = If x > 5 then (x + 2) = (x − 5)

This is again an invalid equation.

Thus, x = satisfies the given equation.

(vi) If x < 1 then x = (x + 1)

x = x + 1

This is an invalid equation.

If 1 < x < 0 then x = (x + 1)

⇒ −x = x + 1

x + x = 1

2x = 1

x = If x > 0 then x = (x + 1)

This is again an invalid equation.

Thus, x = satisfies the given equation.

#### Question 2:

Prove that if 1 < x < 4 and 1 < y < 4, then |x − y| < 3.

Given: 1 < x < 4 and 1 < y < 4

To prove: Proof:

Consider the following cases.

Case 1: x < y

As 1 < x < 4 and 1 < y < 4

1 < x < y < 4

1 x < x x < y x < 4 x (As x is a positive number)

1 x < 0 < y x < 4 x …(1)

Consider,  = < 4 − x (From (1))

< 3 (As 1 < x < 4)

Case 2: x > y

As 1 < x < 4 and 1 < y < 4

4 > x > y > 1

4 y > x y > y y > 1 y (As y is a positive number)

4 y > x y > 0 > 1 y …(2)

Consider,  < 4 − y

< 3 (As 1 < y < 4)

Thus, if 1 < x < 4 and 1 < y < 4, then < 3

#### Question 3:

Find two numbers x and y for which .

The given equation is valid equation when either both x and y are positive or both x and y are negative.

Let us consider x = 2 and y = 3 Now consider x = 2 and y = 3 Thus, x = 2, y = 3 and x = 2, y = 3 satisfy the given equation.

#### Question 4:

Are there numbers x and y for which ?

Yes, there are numbers which satisfy the condition Let us check the validity of the given inequality for both positive and negative numbers.

Let us consider x = 2 and y = 3 (When magnitude of x < magnitude of y) Let us consider x = 5 and y = 3 (When magnitude of x > magnitude of y) Similarly, when x is a positive and y is a negative number.

Thus, the given inequality is valid when x is positive and y is negative or vice-versa.

#### Question 5:

Are there numbers x and y such that ?

If both x and y are positive numbers or both x and y are negative numbers, then If x is positive and y is negative or vice-versa then 