Question

If the real valued function $f\left(x\right)=x^3+3(a^2-1)x+1$ be invertible, then set of possible real values of $a$ is

• A. $(-\infty ,-1)\cup (1,\infty )$
• B. $(-1,1)$
• C. $[-1,1]$
• D. $(-\infty ,-1]\cup (1,+\infty )$

Answer 1

The correct option is A $(-\infty ,-1)\cup (1,\infty )$

$f\left(x\right)=x^3+3(a^2-1)x+1f^{{}^{\prime }}\left(x\right)=3x^2+3(a^2-1)+0f^{{}^{\prime }}\left(x\right)=3x^2+3(a^2-1)$

If function is invertible it must be a one one function

For a function being one one it must be strictly increasing or strictly decreasing.

$\therefore f^{{}^{\prime }}\left(x\right)>0$ or $f^{{}^{\prime }}\left(x\right)<0$

So there are no real roots of $f^{{}^{\prime }}\left(x\right)=0$

$3x^2+3(a^2-1)=0x^2+(a^2-1)=0\therefore b^2-4ac<00-4\left(1\right)(a^2-1)<0(a^2-1)>0(a-1)(a+1)<0$

using number line method

$a\in (-\infty ,-1)\cup (1,\infty )$