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\[\dfrac{1}{2}\left( 2x-1 \right)\le 2x+\dfrac{1}{2}\le 5\dfrac{1}{2}+x\] . Also, find the greatest and the smallest values of x satisfying the inequalities.

(a) \[x\in \left[ -1,4 \right]\]

(b) \[x\in \left[ -1,5 \right]\]

(c) \[x\in \left[ -2,5 \right]\]

(d) \[x\in \left[ -2,7 \right]\]

Answer
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Hint: In this question, we first need to look into the basics of sets and algebra. Then solve the equation like a normal algebraic linear equation but we need to include the inequality. So, that instead of getting a single value of x we get a set of values.

Complete step-by-step answer:

Let us now look at what is meant by set and some basic definitions of algebra.

SET: Set is a collection of well defined objects which are distinct from each other. Sets are usually denoted by capital letters and elements in a set are denoted by small letters.

If a is an element of the set A, then we write it as\[a\in A\]and say a belongs to A or a is in A or a is a member of A.

If a does not belong to A, then we write it as\[a\notin A\].

Roster / Listing Method / Tabular Form: In this method, a set is described by listing elements, separated by commas, within braces.

Now, from the given question let us consider the first part of inequality.

\[\begin{align}

& \Rightarrow \dfrac{1}{2}\left( 2x-1 \right)\le 2x+\dfrac{1}{2} \\

& \Rightarrow \dfrac{1}{2}\left( 2x-1 \right)\le \dfrac{1}{2}\left( 4x+1 \right) \\

& \Rightarrow 2x-1\le 4x+1 \\

\end{align}\]

Now, by rearranging the terms we get,

\[\begin{align}

& \Rightarrow -1-1\le 4x-2x \\

& \Rightarrow -2\le 2x \\

& \therefore x\ge -1 \\

\end{align}\]

Let us consider the other part of the given equation.

\[\Rightarrow 2x+\dfrac{1}{2}\le 5\dfrac{1}{2}+x\]

\[\Rightarrow 2x+\dfrac{1}{2}\le \dfrac{11}{2}+x\]

Now, on rearranging the terms in the above equation we get,

\[\begin{align}

& \Rightarrow 2x-x\le \dfrac{11}{2}-\dfrac{1}{2} \\

& \Rightarrow x\le \dfrac{10}{2} \\

& \therefore x\le 5 \\

\end{align}\]

Here, the minimum value of x is -1 and maximum value of x is 5

Therefore, \[x\in \left[ -1,5 \right]\]

Hence, the correct option is (b).

Note:While considering the inequalities in the given equation it is important to note that on rearranging the terms on the left and right side the inequality remains the same unless it is multiplied with a negative sign.

Here, when we write the set of values of the x we need to note that the minimum value of x is -1 and the maximum value of the x is 5 according to the inequalities obtained in both the parts. So, we need to include all the values between the minimum value of x and the maximum value of x because of the inequality given and write it in the form of a set.

Complete step-by-step answer:

Let us now look at what is meant by set and some basic definitions of algebra.

SET: Set is a collection of well defined objects which are distinct from each other. Sets are usually denoted by capital letters and elements in a set are denoted by small letters.

If a is an element of the set A, then we write it as\[a\in A\]and say a belongs to A or a is in A or a is a member of A.

If a does not belong to A, then we write it as\[a\notin A\].

Roster / Listing Method / Tabular Form: In this method, a set is described by listing elements, separated by commas, within braces.

Now, from the given question let us consider the first part of inequality.

\[\begin{align}

& \Rightarrow \dfrac{1}{2}\left( 2x-1 \right)\le 2x+\dfrac{1}{2} \\

& \Rightarrow \dfrac{1}{2}\left( 2x-1 \right)\le \dfrac{1}{2}\left( 4x+1 \right) \\

& \Rightarrow 2x-1\le 4x+1 \\

\end{align}\]

Now, by rearranging the terms we get,

\[\begin{align}

& \Rightarrow -1-1\le 4x-2x \\

& \Rightarrow -2\le 2x \\

& \therefore x\ge -1 \\

\end{align}\]

Let us consider the other part of the given equation.

\[\Rightarrow 2x+\dfrac{1}{2}\le 5\dfrac{1}{2}+x\]

\[\Rightarrow 2x+\dfrac{1}{2}\le \dfrac{11}{2}+x\]

Now, on rearranging the terms in the above equation we get,

\[\begin{align}

& \Rightarrow 2x-x\le \dfrac{11}{2}-\dfrac{1}{2} \\

& \Rightarrow x\le \dfrac{10}{2} \\

& \therefore x\le 5 \\

\end{align}\]

Here, the minimum value of x is -1 and maximum value of x is 5

Therefore, \[x\in \left[ -1,5 \right]\]

Hence, the correct option is (b).

Note:While considering the inequalities in the given equation it is important to note that on rearranging the terms on the left and right side the inequality remains the same unless it is multiplied with a negative sign.

Here, when we write the set of values of the x we need to note that the minimum value of x is -1 and the maximum value of the x is 5 according to the inequalities obtained in both the parts. So, we need to include all the values between the minimum value of x and the maximum value of x because of the inequality given and write it in the form of a set.