The question provides the following information:
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We are given 4 numbers 1, √2, x and x^2.
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x is a positive number.
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Range of values, i.e., the difference between the highest and the lowest number among the 4 numbers is 4. So:
highest value of List L - lowest value of list L = 4
Try the question with above information once again, if not you can refer to the solution below:
Since square a fraction that is less than one is lesser than the fraction, we can rule out the possibility of x being less than 1, because range is 4. So it is safe to assume 1 is the lowest number out of the 4. so highest value would be x^2 and lowest value would be 1.
x^2 - 1 = 4
x^2 = 5
x = √5
Thus, x is greater than 2.
From x^{2}>x, \sqrt{2} >1, and the range is 4, we could come to conclusion which x^{2}>\sqrt{2} (because the range of L is 4). Hence, the largest number in this series is x^{2}.
Next, to determine which one of the left three numbers is the smallest, three possibilities here:
\square x^{2} > \sqrt{2} >1>x
\blacksquare x^{2}>x>\sqrt{2} >1
\square x^{2}>\sqrt{2}>x>1
Case 1: When x is the smallest, we have x^{2}-x =4,\rightarrow x= \frac{1+\sqrt{17}}{2}>1, so case 1 is out;
Case 2: When 1 is the smallest, we have x^{2}-1=4,\rightarrow x=\sqrt{5}, so Case 2 is true. Then we use \sqrt{5}>\sqrt{2} to rule out Case 3.
In conclusion, the only possible value of x is \sqrt{5}.