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 :
Correct Option: 2,
Solution:
$f(1)=f(2)$
$\Rightarrow 1-a+b-4=8-4 a+2 b-4$
$3 a-b=7$
$f^{\prime}(x)=3 x^{2}-2 a x+b$
$\Rightarrow f^{\prime}\left(\frac{4}{3}\right)=0 \Rightarrow 3 \times \frac{16}{9}-\frac{8}{3} a+b=0$
$\Rightarrow-8 a+3 b=-16$
$a=5, b=8$
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