Question

the number of pairs (a,b) of positive real numbers satisfying a4+b4< 1 and a2+b2>1 is

a) 0 b) 1
c) 2 d) more than 2

Answer 1

the number of pairs (a,b) of positive real numbers satisfying a4+b4< 1 and a2+b2>1 is more than 2

Step-by-step explanation:

a4+b4< 1 and a2+b2>1

as a4+b4< 1

so a & b <1

a4+b4< 1

as AM > GM

a4+b4 > 2a2b2

2a2b2 < 1

a2+b2>1

(a - b)2 + 2a2b2 > 1

as 2a2b2 < 1

so both cases possible satisfying and non satisfying

let s take few example

4/6 5/6

a2+b2 = 16/36 + 25/36 = 41/36 > 1

a4+b4 = 256/1296 + 625/1296 = 881/1296< 1

5/7 , 6/7

a2+b2 = 25/49 + 36/49 = 61/49 > 1

a4+b4 = 625/2401 + 1296/2401 = 1921/2401 < 1

6/8 , 7/8

a2+b2 = 36/64 + 49/64 = 85/64 > 1

a4+b4 = 1296/4096 + 2401/4096 = 3697/4906 < 1

hence we can say more than 2 solutions exist.