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f(a)=b on when a>3\(\sqrt{5}\)

a=\(\frac{(b^3+20)}{b}\).........Eq-1

Ques : Find m,n so that f(m) = n

m cannot be less than 3\(\sqrt{5}\), so m can only be 5, 14, 15, 21
As there are no restrictions on n we can put values in the RHS of Eq1 (replace b with n), and find m

When n is:
2, \(\frac{(n^3+20)}{n}\) = \(\frac{(8+20)}{2}\) = 14 (corresponding value of m, even though we have got the answer I will calculate further, just for clarity or may be fun ;) )

3, \(\frac{(n^3+20)}{n}\) = \(\frac{(27+20)}{3}\) = \(\frac{47}{3}\) (corresponding value of m)

5, \(\frac{(n^3+20)}{n}\) = \(\frac{(125+20)}{5}\) = \(29\) (corresponding value of m)

We do not need to calculate further as the value of m will only keep increasing.

m=14 & n=2
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This is an Easy-Medium level question, everyone got correct. OA has been added. Last couple of questions will be posted tomorrow. Good Luck for today's questions.

I will update the points table shortly.

Good Luck
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f(a)=b on when a>3\(\sqrt{5}\)

a=\(\frac{(b^3+20)}{b}\).........Eq-1

Ques : Find m,n so that f(m) = n

m cannot be less than 3\(\sqrt{5}\), so m can only be 5, 14, 15, 21
As there are no restrictions on n we can put values in the RHS of Eq1 (replace b with n), and find m

When n is:
2, \(\frac{(n^3+20)}{n}\) = \(\frac{(8+20)}{2}\) = 14;) )

3, \(\frac{(n^3+20)}{n}\) = \(\frac{(27+20)}{3}\) = \(\frac{47}{3}\) (corresponding value of m)

5, \(\frac{(n^3+20)}{n}\) = \(\frac{(125+20)}{5}\) = \(29\) (corresponding value of m)

We do not need to calculate further as the value of m will only keep increasing.

m=14 & n=2
­if m cant be less that 3\(\sqrt{5}\), then m can only be 14,15,& 21. we should eliminate 5 as well, right? 
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