If $2$ and $3$ are the roots of cubic equation $2x^3+mx^2-13x+n=0$, then the values of $m$ and $n$ are:
- $-5,-30$
- $-5,30$
- $5,30$
- $None\ of\ these$
If 2 and 3 are roots, then (x-2) and (x-3) are factors. The third root r can be found using the product of roots: 2 * 3 * r = -n/2, and sum of roots 2 + 3 + r = -m/2. Using the coefficient of x, the sum of roots taken two at a time is (2*3 + 2*r + 3*r) = -13/2, which gives 6 + 5r = -6.5, so 5r = -12.5 and r = -2.5. Then m = -2*(2+3-2.5) = -5 and n = -2*(2*3*-2.5) = 30.
Since 2 and 3 are roots of the cubic equation 2x^3 + mx^2 - 13x + n = 0, we substitute them directly into the equation to form a system. For x = 2, the equation is 16 + 4m - 26 + n = 0, which simplifies to 4m + n = 10. For x = 3, the equation is 54 + 9m - 39 + n = 0, which simplifies to 9m + n = -15. Subtracting the first equation from the second gives 5m = -25, so m = -5. Substituting m back into the first equation yields -20 + n = 10, so n = 30. Therefore, the values are m = -5 and n = 30.