Short Answer Questions 1 — Question 4

(a) Given,

$\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x$

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

$\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \ge -1 \text{ ...........(1)}$

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

$\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \\[1em] \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \\[1em] \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \le \dfrac{10}{2} \\[1em] \Rightarrow x \le 5 \text{ ...........(2)}$

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.