For each value of a 3 (5^1/2) the function f(a) is such that the

Question

For each value of  a > 3\sqrt{5} , the function  f(a)  is such that the equation  f(a) = b  has a solution of the form  a = \frac{b^3 + 20}{b}

Select one value for  m  and one value for  n  such that the information implies that  f(m) = n , based on the defined function. Make only two selections, one in each column.

Deepakjhamb · Answer

Select one value for m and one value for n such that the information implies that f(m)=n, based on the defined function. Make only two selections, one in each column.

m n 
2
3
5
14
15
21

solution :

first of all lets check constraints , we are given that a>3* underoot 5 means a>6.7

so now m represents a and n represents b , also we need to find a and b such that a = (b^3 +20)/b

so first of all m cannot be equal to 2 ,3 or 5 as a> 6.7

so lets plugin a = 14 from the solution set and do hit and trial , from this we get 14 = (b^3 +20 )/b so only 2 as value of b or n satisfies this equation as (2^3 +20)/2 = 28/2 = 14

lets take other values for b^3+ 20/b if we put b = 3 we get 47/3 which is not even an integer that are part of solution set

lets take b=5 , then b^3 +20/b = 145/5 = 29 which is not part of solution set as largest integer is 21

now if we take (14^3 +20)/14 we will not get an integer so out

now if we take 15^3 + 20 /15 will also not give integer so out

also 21^3 + 20 /20 will not give integer so out .

also one inference we could have used was that b has to be either multiple of 2 or 5 to get an integer out of (b^3 +20 )/b as we cannot add to a multiple of number a non multiple and then divide by same number to get an integer

so clearly from above solution set for this question reduces to m = 14 and n =2

Sumi1010 · Answer

The solution of the equation f(a) = b has the form a =  /b where a > 3 sq root(5)

Thus, the solution to for same function f(m) = n will have the relation

m =   / n for m > 3 sq root(5)
i.e.  m = n^2 + (20/n)

For n = 2, m comes out to 14 
For n = 3, m comes out to 47/3
For n = 5, m comes out to 29
For n = 14, m comes out to 1382/7
For n = 15, m comes out to 225(4/3)
For n = 21, m comes out to 441(20/21)

Thus,  n = 2, m = 14

rocky620 · Answer

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

Sajjad1994 · Answer

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

033 · Answer

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

For each value of a > 3 (5^1/2) the function f(a) is such that the

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