Question

The function $x\left( x-1 \right)\left( x-2 \right)$ attains its maximum value when the value of x is equal to
(a) 1
(b) $1+\dfrac{1}{\sqrt{3}}$
(c) $1-\dfrac{1}{\sqrt{3}}$
(d) $1\pm \sqrt{3}$

Answer 1

Hint:We will first simplify $x\left( x-1 \right)\left( x-2 \right)$ and then find its derivative with respect to x. In the final step, we will find the value of x by equating the obtained derivative equal to 0. If $x<0$ then it is a point of maxima, and if $x>0$ then we have the point of minima.

Complete step-by-step answer:
It is given in the question that we have to find the value of x for which the given expression $x\left( x-1 \right)\left( x-2 \right)$ results in its maximum value. Let us assume that $f\left( x \right)=x\left( x-1 \right)\left( x-2 \right)$. Simplifying further we get,
$f\left( x \right)=\left( {{x}^{2}}-x \right)\left( x-2 \right)$
Opening the brackets to get a cubic function of x, that is
$f\left( x \right)={{x}^{3}}-2{{x}^{2}}-{{x}^{2}}+2x$
Finally, we get,
$f\left( x \right)={{x}^{3}}-3{{x}^{2}}+2x$
Now, we will find the derivative of function $f\left( x \right)$ with respect to x. That is,
$\dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d{{x}^{3}}}{dx}-\dfrac{d\left( 3{{x}^{2}} \right)}{dx}+\dfrac{d\left( 2x \right)}{dx}$.
Now, we know the following formula for derivative as $\dfrac{d{{\left( x \right)}^{n}}}{dx}=n{{x}^{n-1}}$. Using this formula in the derivative of $f\left( x \right)$ we get ${{f}^{'}}\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}=3{{x}^{2}}-6x+2$.
For finding the derivative of $f\left( x \right)$ we will equate its derivative $f'\left( x \right)$ to 0. That is
${{f}^{'}}\left( x \right)=3{{x}^{2}}-6x+2=0$ therefore we require the roots of this equation which are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
On putting the values in this, we get
$x=\dfrac{6\pm \sqrt{{{6}^{2}}-4\times 3\times 2}}{2\times 3}$
Solving further, we get
$x=\dfrac{6\pm \sqrt{36-24}}{6}=\dfrac{6\pm \sqrt{12}}{6}$
Simplifying further by multiplying and dividing by $\sqrt{3}$ in the second term of equation, we get
$x=\dfrac{6}{6}\pm \dfrac{\sqrt{12}\times \sqrt{3}}{6\times \sqrt{3}}=1\pm \dfrac{\sqrt{36}}{6\times \sqrt{3}}$, that is, we get the value of x as
$x=1\pm \dfrac{6}{6\sqrt{3}}=1\pm \dfrac{1}{\sqrt{3}}$.
Now, we have two values of x as $x=1+\dfrac{1}{\sqrt{3}}$ and $x=1-\dfrac{1}{\sqrt{3}}$. We know that if $x<0$ then it is a point of maxima and if $x>0$ then it is a point of minima.
Since $x=1+\dfrac{1}{\sqrt{3}}>0$ therefore it is a point of minima and $x=1-\dfrac{1}{\sqrt{3}}<0$ thus it is a point of maxima. Thus we get option c) as the correct answer.

Note: Student may get confused in the last part of solution that’s why we are considering only $x=1-\dfrac{1}{\sqrt{3}}$ as the point of maxima, if they do not know that concept that when $x<0$ then it is a point of maxima. As a result they may tick the wrong option even after solving the whole question correctly.Students should remember the derivatives formulas of polynomial functions to solve these types of questions.