Question

If $t$ is a real number satisfying the equation $2t^3-9t^2+30-a=0,$ the find the value of the parameter a for which the equation $x+\frac{1}{x}=t$ gives six real and distinct values of $x.$

Answer 1

$2t^3-9t^2+30-a=0$

$f\left(t\right)=2t^3-9t^2+30-a$

$f\left(t\right)=6t^2-18t$

$f\left(t\right)=0$

$6t(t-3)=0$

$t=0,$ $t=3$

$f\left(0\right),f\left(3\right)<0$

$(30-a).(54-81+30-a)<0$

$(30-a)(3-a)<0$

$a\in (3,30)\to \left(i\right)$

$\Rightarrow x+\frac{1}{x}=t,\left|x\right|⩾2$

No roots lie between $[-2,2]$

$\Rightarrow f(-2)>0,$ and $f\left(2\right)>0$

$\Rightarrow -16+30-36-a>0$

$a<-22$

Also

$\Rightarrow 16-36+30-a>0$

$a<10$

$a<-22\to \left(ii\right)$

= there is no real value of a.