Nikkb wrote:
b is an integer
value of :\(\frac{4b + 1}{2b}\) when rounded to the nearestinteger
\(\frac{4b + 1}{2b} = 2+\frac{1}{2b}\)
=>for b =+ve integer => maximum value of\(\frac{1}{2b}\) will be for b=1 =>\(2 + \frac{1}{2} = 2+0.5 = 2.5 = 3\) rounded to nearest integer
=> for\(b >1\) value of\(\frac{1}{2b}\) will be less than 0.5 => \(2 + \frac{1}{2*2} = 2+0.25 =2.25 = 2\) rounded to nearest integer
so for all values of b > 1 nearest integer value for\(\frac{4b + 1}{2b}\) = 2
=> for b=0 value
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