Mathematics Part ii Solutions Solutions for Class 9 Math Chapter 4 - Real Numbers

Answer:

(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:

Answer:

(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.

Answer:

(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.

Answer:

(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.

Answer:

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

Answer:

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.

Answer:

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.

Answer:

No, the given inequality is not valid for any number x and y.

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