Question

The equation $x^3-3x+\left[a\right]=0$ will have three real and distinct roots, then the set of all possible values of $a$ is
(where $[\cdot ]$ denotes the greatest integer function)

• A. $(-\infty ,-2)\cup (2,\infty )$
• B. $[-1,2)$
• C. $[2,\infty )$
• D. $(-2,-1]$

Answer 1

The correct option is B $[-1,2)$

Let $f\left(x\right)=x^3-3x$ and $g\left(x\right)=-\left[a\right]$
And the equation $f\left(x\right)=g\left(x\right)$ has three real and distinct roots$\cdots \left(i\right)$
For, $f\left(x\right)=x^3-3x$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-3=3(x-1)(x+1)$
$\Rightarrow f^{\prime }\left(x\right)=0$
$\Rightarrow x=\pm 1$ are critical points and $f^{\prime \prime }\left(x\right)=6x$
$\Rightarrow f^{\prime \prime }(-1)=-6$
$\Rightarrow$ at $x=-1$ (local maxima)
$\Rightarrow f^{\prime \prime }\left(1\right)=6$
$\Rightarrow$ at $x=1$ (Local minima)
$f\left(0\right)=0f(-\infty )\to -\infty f\left(\infty \right)\to \infty$
Now, plotting graph of $f\left(x\right):$

For equation $\left(i\right),$
$\Rightarrow -2<g\left(x\right)<2$
$\Rightarrow -2<-\left[a\right]<2$
$\Rightarrow -2<\left[a\right]<2$
$\therefore a\in [-1,2)$