Rolle's theorem holds for the function $f(x)=x^3+ax^2+\beta x, 1≤x≤2$ at the point $\frac{4}{3}$, the values of $\alpha $ and $\beta $ are :

$\alpha=-5, \beta =8$

The correct answer is option (4) → $\alpha=-5, \beta =8$

$f(x)=x^3+\alpha x^2+\beta x$

$f'(x)=3x^2+2\alpha x+\beta =0$ (By rolle's theorom)

at $\frac{4}{3}$ ⇒ $\frac{16}{3}+\frac{2\alpha x4}{3}+\beta =0$

$⇒16+8α+3β=0$   ...(1)

also $f(1)=f(2)$

$⇒1+α+β=8+4α+2β$

so $3α+β+7=0$   ...(2)

eq. (1) - 3 eq. (2)

$16+8α+3β=0-(21+9α+3=0)$

$α=-5⇒β=8$