Question

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

• A. $(-5,8)$
• B. $(5,8)$
• C. $(5,-8)$
• D. $(-5,-8)$

Answer 1

The correct option is B $(5,8)$

$f\left(1\right)=f\left(2\right)$
$\Rightarrow 1-a+b-4=8-4a+2b-4$
$3a-b=7$ $\dots \left(1\right)$
$f^{\prime }\left(x\right)=3x^2-2ax+b$
$\Rightarrow f^{\prime }\left(\frac{4}{3}\right)=0\Rightarrow 3\times \frac{16}{9}-\frac{8}{3}a+b=0$
$\Rightarrow -8a+3b=-16\dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$a=5,b=8$