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 :
Answer & explanation
Correct answer: option 4
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)
$16+8α+3β=0-(21+9α+3=0)$
$α=-5⇒β=8$