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 $f^{\text{'}}\left(\frac{4}{3}\right)=0$, then ordered pair $\left(a,b\right)$ is equal to

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

Answer 1

The correct option is B

$(5,8)$

Finding the ordered pair $\left(a,b\right)$:

Step 1: Rolle's theorem states that:

If a function $f$ is defined in the closed interval $a,b$ in such a way that it meets the conditions below:

On the closed interval $a,b$, the function $f$ is continuous.

On the open interval, the function $f$ is differentiable $\left(a,b\right)$

If $f\left(a\right)=f\left(b\right)$, then at least one value of $x$ exists; let us assume that this value is $c$, which is between $a$ and $b$, in such a way that $f\text{'}\left(c\right)=0$.

Here the function $f\left(x\right)=x^3-a{x}^2+bx-4$ which is defined on $x\in [1,2]$ satisfies Rolle’s theorem, then

$$\begin{gathered} \therefore f\left(1\right)=f\left(2\right) \\ \begin{array}{l}\Rightarrow {\left(1\right)}^3-a{\left(1\right)}^2+b\left(1\right)-4={\left(2\right)}^3-a{\left(2\right)}^2+b\left(2\right)-4\\ \Rightarrow 1-a+b=8-4a+2b\\ \Rightarrow -a+b+4a-2b=8-1\\ \Rightarrow 3a-b=7..............\left(i\right)\end{array} \end{gathered}$$

Step 2: Differentiate $f\left(x\right)=x^3-a{x}^2+bx-4$ with respect to $x$ and substitute $x=\frac{4}{3}$

$\begin{array}{l}\therefore f^{\text{'}}\left(x\right)=3x^2-2ax+b\\ \mathrm{given}{f}^{\text{'}}\left(\frac{4}{3}\right)=0\\ \Rightarrow 3{\left(\frac{4}{3}\right)}^2-2a\left(\frac{4}{3}\right)+b=0\\ \Rightarrow \frac{16}{3}-\frac{8a}{3}+b=0\\ \Rightarrow 16-8a+3b=0\\ \Rightarrow 8a-3b=16.............\left(ii\right)\end{array}$

Step 3: Solve equations $\left(i\right)$ and $\left(ii\right)$

Multiply equation $\left(i\right)$by $3$,

$9a-3b=21........\left(iii\right)$

Then subtract equation $\left(ii\right)$from the equation $\left(iii\right),$ we get $a=5$

Now substitute $a=5$ in equation $\left(i\right)$, then

$\begin{array}{l}\Rightarrow 3\left(5\right)-b=7\\ \Rightarrow 15-7=b\\ \Rightarrow b=8\end{array}$

Therefore, ordered pair $\left(a,b\right)$ is equal to $\left(5,8\right)$.

Hence, the correct option is (B).