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Given $\displaystyle \cfrac{\;\cfrac{a}{b} + \cfrac{b}{a}\;}{\cfrac{a}{b} - \cfrac{b}{a}} = 1$

If $a$ and $c$ are positive integers, then how many ordered pairs are possible for $(a,c)$, where: $a + 4b^2 + c \leq 8$?

  1. 45
  2. 28
  3. 17
  4. 18

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3 Answers

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3 votes
3 votes

Ans will be 28, 

((a^2+b^2)/ab)/((a^2-b^2)/ab)=1

=>(a^2+b^2)/(a^2-b^2)=1

=>(a^2+b^2)=(a^2-b^2)

=>2b^2=0

=>b=0

a+c<=8   Because a and c are +ve integers.

=>a=7          c=1                         =1 ordered pair

=>a=6         c=1,2                      =2 ordered pair    

...............

=>a=1          c=1,2.......,7           =7 ordered pair

no, of ordered pair 1+2+.......7= 28 

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3 votes
3 votes

Ans will be 45

((a^2+b^2)/ab)/((a^2-b^2)/ab)=1

=>(a^2+b^2)/(a^2-b^2)=1

=>(a^2+b^2)=(a^2-b^2)

=>2b^2=0

=>b=0

a+c<=8

=>a=8          c=0                            =1 ordered pair

=>a=7          c=0,1                         =2 ordered pair

=>a=6         c=0,1,2

...............

=>a=0          c=0,1,2.......,8               =9 ordered pair

no, of ordered pair 1+2+..........9=45 

0 votes
0 votes
This is an easy question.

after getting b=0, we'll have to do little observation.

suppose i ask you to find how many pairs give sum equal to 2. answer is 1 way

again, how many pairs give sum equal to 3, answer is 2 ways

how many pairs give sum equal to 4 , answer is 3 ways.

similarly, how many pairs give sum equal to 8 , answer is 7 ways.

therefore, it is nothing but the sum of first 7 natural number starting from 1 which gives (7*8)/2= 28. answer.

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