Question

The number of pairs (a, b) of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$ is

• A. $0$
• B. $1$
• C. $2$
• D. More than $2$

Answer 1

The correct option is D More than $2$

$a^4+b^4<1$

$a^2+b^2>1$

$-a^2-b^2<-1$

$a^4-a^2+b^4-b^2<0$

$(a^2-\frac{1}{2}{)}^2+(b^2-\frac{1}{2}{)}^2<\frac{1}{2}$

Hence, it is a circle of radius $\frac{1}{\sqrt{2}}$ and all the points within it satisfies this equation. Hence, more than 2