Question

If f($x+\frac{1}{x}$) = $\frac{x^3+1}{x^3}$ (x $\ne$ 0 ) then

• A. f(x) is increasing function
• B. f(x) has a local maximum at x= -1
• C. f(x) is injective in its domain of definition
• D. The equation f (x) = 3 has a unique real root

Answer 1

Answer: (A), (C), (D)

The correct options are
A

f(x) is increasing function

C

f(x) is injective in its domain of definition

D

The equation f (x) = 3 has a unique real root

$f(x+\frac{1}{x}$) = $x^3$ + $\frac{1}{x^3}$ = ${(x+\frac{1}{x})}^3$ - $3x\frac{1}{x}$$(x+\frac{1}{x}$)

$f\left(x\right)=x^3-3x$

Since $x+\frac{1}{x}$ $\ge$ 2 for$x$ $>$ 0 and $x+\frac{1}{x}$ $\le$ - 2,for $x<0$, the domain of

$f\left(x\right)is(-\infty$ , -2) $\cup$ [ 2 , $\infty$ ]

$f^{{}^{\prime }}\left(x\right)$ = 3 ($x^2$ - 1) $>$ 0 $\forall x$ $\in$ D

Thus $f\left(x\right)$ is increasing in D , injective there and $f\left(s\right)=3$ has a unique real root.

Since f is not defined at $x$ = -1`, (b) does not hold