A real number x when added to its reciprocal give minimum value to the...
Finding the Real Number that Gives Minimum Value to the Sum
Solution:
Let x be the real number. Then, its reciprocal is 1/x.
The sum of x and its reciprocal is x + 1/x.
To find the minimum value of this sum, we can use the concept of the arithmetic mean and geometric mean inequality.
We know that for any two positive numbers a and b, the arithmetic mean is (a+b)/2 and the geometric mean is √(ab).
The arithmetic mean is always greater than or equal to the geometric mean, i.e., (a+b)/2 ≥ √(ab).
Let's apply this inequality to x and 1/x.
The arithmetic mean of x and 1/x is (x + 1/x)/2.
The geometric mean of x and 1/x is √(x * 1/x) = √1 = 1.
By the arithmetic mean and geometric mean inequality, we have:
(x + 1/x)/2 ≥ √(x * 1/x) = 1
Multiplying both sides by 2 gives:
x + 1/x ≥ 2
Therefore, the minimum value of x + 1/x is 2, which is attained when x=1.
Hence, the real number x that gives minimum value to the sum x + 1/x is 1.
A real number x when added to its reciprocal give minimum value to the...
When x is 1:x + 1/x 》 1+ 1/1 = 2When x is more than 1(i.e. 2)x + 1/x 》 2 + 1/2 = 5/2 = 2.5And the result will keep on increasing when x = 3,4,5,.....nWhen x is less than 1(i.e. 1/2)x + 1/x 》 1/2 + 2 = 5/2 = 2.5And the result will keep on increasing when x = 1/3, 1/4, 1/5,.....n or any no. Less than 1Therefore, minimum value of the sum of a real no. And its reciprocal is derived when x = 1