Question

If x E{-2,-1,0,1,2,3,4,5}, find the solution set of each of the following inequations: (i) 3x + 4 < 15 (i) 2/3 +x - 1/6


Kardona solve

Answer 1

Explanation:

Explanation:

Explanation:

We are given two inequalities which has the solution that belongs to { -2, -1, 0, 1, 2, 3, 4, 5 }

Inequality (i)

⇒ 3x + 4 < 15

⇒ 3x < 11

⇒ x < 11/3

⇒ x < 3.666...

Which means every value of x which is less than 3.6 satisfies the given inequality, but it is given in the question that x must belong to the given set of solutions.

So, We have the following solution

⇒ x ∈ { -2, -1, 0, 1, 2, 3 }

Inequality (ii)

⇒ 2/3 + x < 1/6

⇒ x < 1/6 - 2/3

⇒ x < (1 - 4)/6

⇒ x < -3/6

⇒ x < -0.5

Which means every value of x less than -0.5 satisfies the given inequality. But x must belong to { -2, -1, 0, 1, 2, 3, 4, 5 }, so we have

⇒ x ∈ {-2, -1}

Some Information :-

☞ A set is a collection of objects or item, that are of the same type, like Set of Integers, Set of all Cities, Set of all countries, e.t.c

☞ The number of Subsets of a set can be given by 2^{n} where n is the number of elements in the set.

Answer 2

QUESTION:

Find the domain and range of the real function f denoted by f(x) = √x - 1

GIVEN:

f(x) = √x - 1

TO FIND:

Domain

Range

SOLUTION:

Firstly finding the Domain

f(x) = √x - 1

Seperate the function into two parts

√x - 1

x - 1

Find all values for which the radicand is positive or 0 Domain is all real numbers

x ≥ 1

x belongs to R

Where R is real numbers

Find the union

x belongs to [1 , ∞)

Alternative form

x ≥ 1

Now, finding the range

Let

y = √x - 1

→ y² = x - 1

→ y² + 1 = x

→

y² ≥ 0

Now,

y belongs to [0 , ∞)

Refer the attachment