If `t` is a real number satisfying the equation `2t^3-9t^2+30-a=0,` then find the values of the parameter `a` for which the equation `x+1/x=t` gives six real and distinct values of `x` .

Explanation

we have `2t^(3)-9r^(2)+30-a=0`
Now `t=x+(1)/(x)`
Since `|x+(1)/(x)|ge2|t|ge2`
So any real root `t_(0) ` of equation (1)gives two real and distinct values of x if `|t_(0)|gt2`
Thus we need to find the condition for the equation in 't' to have three real and distinct roots none of which lies in [-2,2]
Let `f(t)=2t^(3)-9t^(2)+30-a`
`f(t)=6t^(2)-18t=0`
6=0,3

So the equation f(t)=0 has three real and distinct roots if `f(x).f(3)lt0`
`(30-a)(54-81+30-a)lt0`
`(30-a)(3-a)lt0`
`a in(3,30)`
Also none of the roots lies in [-2,2] if `(-2)gt "and" f(2)gt0`
`-16=36+30-a gt 0 "and" 16-36+30-a gt 0`
`-22-a gt 0 "and" 10-a gt 0`
`a+22 lt 0 "and" a -10 lt 0`
`alt-22 "and" alt10`
`alt-22`
From (1) and (2) no real value of a exits