For which of the following values of real number $t$, the equation $x^4-tx+\dfrac 1t = 0$ has no root on the interval $[1,2]$.

Question

For which of the following values of real number $t$, the equation $x^4-tx+\dfrac 1t = 0$ has no root on the interval $[1,2]$. I think we might check for which $t$ $y=tx-\dfrac 1t $ have solution on [1,16].