Question:
If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $\mathrm{f}^{\prime}\left(\frac{4}{3}\right)=0$, then ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to $:$
Correct Option: 1
Solution:
$f(1)=f(2)$
$\Rightarrow 1-a+b-4=8-4 a+2 b-4$
$\Rightarrow 3 a-b=7$ $\ldots \ldots .(1)$
Also $\mathrm{f}^{1}\left(\frac{4}{3}\right)=0$ (given)
$\Rightarrow\left(3 x^{2}-2 a x+b\right)_{x-\frac{4}{3}}=0$
$\Rightarrow \frac{16}{3}-\frac{8 a}{3}+b=0$
$\Rightarrow 8 a-3 b-16=0$ ...........(2)
Solving (1) and (2)
$a=5, b=8$
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