Question

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

Answer 1

Hint: Find the derivative of the given function with respect to $x$ and then solve the equation $f'\left( x \right) = 0$. Since $f\left( x \right)$ is a function of degree $3$, so the degree of it’s derivative will be $2$. Hence solving $f'\left( x \right) = 0$, you will get two values of $x$. You need to take the value of $x$ which lies between $1$ and $2$.

Formula Used:
$\dfrac{d}{{dx}}\left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} \pm \dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}$, where $f\left( x \right)$ and $g\left( x \right)$ are two functions of $x$.
$\dfrac{d}{{dx}}\left\{ {cf\left( x \right)} \right\} = c\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$, where $c$ is a constant
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, where $n$ is a real number
and the derivative of constant is zero.

Complete step by step solution:
Given that $f\left( x \right) = {x^3} - 3{x^2} + 2x + 10$
Differentiating with respect to $x$, we get $f'\left( x \right) = 3{x^2} - 6x + 2$
Now, solve the equation $f'\left( x \right) = 0$
$\therefore 3{x^2} - 6x + 2 = 0$
This is a quadratic equation.
Comparing this equation with the general form $a{x^2} + bx + c = 0$, we get $a = 3,b = - 6,c = 2$
By Shridharachariya’s law, we get
$x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4 \times 3 \times 2} }}{{2 \times 3}}$
$= \dfrac{{6 \pm \sqrt {36 - 24} }}{6}\\ = \dfrac{{6 \pm \sqrt {12} }}{6}\\ = \dfrac{{6 \pm 2\sqrt 3 }}{6}\\ = \dfrac{{2\left( {3 \pm \sqrt 3 } \right)}}{6}\\ = \dfrac{{3 \pm \sqrt 3 }}{3}$
The value of $\sqrt 3 $ is approximately $1.732$
So, $x = \dfrac{{3 \pm 1.732}}{3}$
Taking positive sign, we get $x = \dfrac{{3 + 1.732}}{3} = \dfrac{{4.732}}{3} > 1$
Taking negative sign, we get $x = \dfrac{{3 - 1.732}}{3} = \dfrac{{1.268}}{3} < 1$
You have to find such a value of $x$ which lies between $1$ and $2$.
So, take $x = \dfrac{{3 + \sqrt 3 }}{3}$

Option ‘A’ is correct

Note: Here you have to use Sridharachariya’s formula $x = \dfrac{{ - \left( {b} \right) \pm \sqrt {{{\left( {b} \right)}^2} - 4 \times a \times c} }}{{2 \times a}}$. So, you should be aware of this formula and check which value of $x$ lies between $1$ and $2$.