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)-9t^(2)+30-a=0`.
Any real root `'t_(0)'` of this equation gives two real and distinct values of x from `x+(1)/(x)=t` if `|t_(0)| gt 2`.
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 rArr t = 0, 3`

So the equation `f(t)=0` has three real and distinct roots if `f(0)*f(3) lt 0`
`rArr (30-a)(54-81+30-a) lt 0 rArr (30-a)(30-a) lt 0`
`rArr (a-3)(a-30) lt 0 rArr a in (3, 30) " " (i)`
Also, none of the roots lies in [-2,2] if `f(-2) gt 0` and `f(2) gt 0 (.:' -2, 2 in (A, B))`
`-6-36+30-a gt 0 ` and `16 -36+30-a gt0`
`-22-a gt 0 ` and `10-a gt 0 rArr a+22 lt 0 ` and `a-10 lt 0`
`rArr a lt -22` and `a lt 10 `
`rArr a lt -22 " " (ii) `
From (i) and (ii), no real value of a exists.