Question

A value of $x$ in the interval $\left(1,2\right)$ such that $f\text{'}\left(x\right)=0$ where $f\left(x\right)=x^3-3x^2+2x+10$ is

• A. $\frac{3+\sqrt{3}}{3}$
• B. $\frac{3+\sqrt{2}}{2}$
• C. $1+\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

The correct option is A

$\frac{3+\sqrt{3}}{3}$

Explanation for correct answer:

Finding the value of X:

Step-1: Differentiate the given function.

Given, $f\left(x\right)=x^3-3x^2+2x+10$

Differentiate with respect to $x$.

$\Rightarrow$$f\text{'}\left(x\right)=3x^2-3\times 2x+2\times 1+0$

$\Rightarrow$ $0=3x^2-6x+2\left(\because f\text{'}\left(x\right)=0\right)$

Step-2: Finding roots of quadratic equation.

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow$ $x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\times 3\times 2}}{2\times 3}$

$\Rightarrow$ $x=\frac{6\pm 2\sqrt{3}}{6}$

$\Rightarrow$ $x=\frac{3\pm \sqrt{3}}{3}$

Hence, option $\left(A\right)$ is correct.