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Mathematics

Solve the following inequation and answer the questions given below.

12(2x1)2x+12512+x\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x

(a) Write the maximum and minimum values of x for x ∈ R.

(b) What will be the change in maximum and minimum values of x if x ∈ W.

Linear Inequations

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Answer

(a) Given,

12(2x1)2x+12512+x\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x

Solving L.H.S. of the inequation, we get :

12(2x1)2x+12x122x+122xx1212x1 ………..(1)\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \ge -1 \text{ ………..(1)}

Solving R.H.S. of the inequation, we get :

2x+12512+x2x+12112+x2xx11212x102x5 ………..(2)\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \\[1em] \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \\[1em] \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \le \dfrac{10}{2} \\[1em] \Rightarrow x \le 5 \text{ ………..(2)}

From equation (1) and (2), we get :

-1 ≤ x ≤ 5 and x ∈ R.

Hence, minimum and maximum value of x is -1 and 5 respectively.

(b) If x ∈ W.

Then, minimum value = 0 and maximum value = 5.

Hence, minimum and maximum value of x is 0 and 5 respectively, when x is a whole number.

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