Question

Suppose that $a$ and $b$ are nonzero real numbers and if $a<\frac{1}{a}<b<\frac{1}{b}$ then $a<-1$.

• A. True
• B. False

Answer 1

The correct option is A True

Given nonzero reals $a,b$ such that $a<\frac{1}{a}<b<\frac{1}{b}$:

$a\times a^2<\frac{1}{a}\times a^2$.

Thus $a^3<a.$

$\Rightarrow a^3-a<0$

$\Rightarrow a(a^2-1)<0$

$\Rightarrow a(a+1)(a-1)<0$.

$\Rightarrow a<-1$ or $0<a<1.$

$a\times b^2<\frac{1}{b}\times b^2$.

Thus $b^3<b.$

$\Rightarrow b^3-b<0$

$\Rightarrow b(b^2-1)<0$

$\Rightarrow b(b+1)(b-1)<0$.

$\Rightarrow b<-1$ or $0<b<1.$

If $0<a<1\Rightarrow \frac{1}{a}>1>b$

which is a contradiction.

$\therefore a<-1$.