# If Rolle's theorem holds for the function

Question:

If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f^{\prime}\left(\frac{4}{3}\right)=0$, then ordered pair $(a, b)$ is equal to :

1. (1) $(-5,8)$

2. (2) $(5,8)$

3. (3) $(5,-8)$

4. (4) $(-5,-8)$

Correct Option: 2,

Solution:

$f(1)=f(2)$

$\Rightarrow 1-a+b-4=8-4 a+2 b-4$

$3 a-b=7$

$f^{\prime}(x)=3 x^{2}-2 a x+b$

$\Rightarrow f^{\prime}\left(\frac{4}{3}\right)=0 \Rightarrow 3 \times \frac{16}{9}-\frac{8}{3} a+b=0$

$\Rightarrow-8 a+3 b=-16$

$a=5, b=8$