Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The constant c of Lagrange's theorem for $f\left( x \right)=x\left( x-1 \right)\left( x-2 \right)$ in $\left[ 0,\dfrac{1}{2} \right]$ is\begin{align} & A.\dfrac{1}{2} \\ & B.\dfrac{6+\sqrt{21}}{6} \\ & C.\dfrac{6-\sqrt{21}}{6} \\ & D.\dfrac{\sqrt{21}-6}{6} \\ \end{align}

Last updated date: 22nd Jul 2024
Total views: 424.2k
Views today: 13.24k
Verified
424.2k+ views
Hint: To solve the question given above, we will use the concept of Lagrange's mean value theorem (LMVT). So, first we will find out what is LMVT and then using it, we will find the value of $f'\left( c \right)$ as shown below:
$f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$
Then, we will find the derivative of the function $\left( f'\left( x \right) \right)$ and put c in place of it. We will solve the quadratic in c with the help of the quadratic formula.

Complete step by step answer:
Before we solve the question, we must know what Lagrange’s mean value theorem (LMVT) is. Lagrange's mean value theorem (LMVT) states that if a function $f\left( x \right)$ is continuous on closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ then there is at least one point $x=c$ on this interval, such that:
$f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$
In our case, $f\left( x \right)=x\left( x-1 \right)\left( x-2 \right)$ which is continuous and differentiable in the interval $\left[ 0,\dfrac{1}{2} \right]$ .Now, we will calculate $f'\left( c \right)$ first with the help of LMVT. In our case, $a=0\,\,\text{and }b=\dfrac{1}{2}$ Thus, we have:
\begin{align} & f'\left( c \right)=\dfrac{f\left( \dfrac{1}{2} \right)-f\left( 0 \right)}{\dfrac{1}{2}-0} \\ & \Rightarrow f'\left( c \right)=\dfrac{f\left( \dfrac{1}{2} \right)-f\left( 0 \right)}{\left( \dfrac{1}{2} \right)} \\ & \Rightarrow f'\left( c \right)=2\left( f\left( \dfrac{1}{2} \right)-f\left( 0 \right) \right)......................(i) \\ \end{align}
Now, we will calculate the values of $f\left( \dfrac{1}{2} \right)\text{ and }f\left( 0 \right)$ Thus, we have:
\begin{align} & f\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2}-1 \right)\left( \dfrac{1}{2}-2 \right) \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{2}\times \left( \dfrac{-1}{2} \right)\left( \dfrac{-3}{2} \right) \\ & \Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{3}{8}\,.........................(ii) \\ \end{align}
\begin{align} & f\left( 0 \right)=0\left( 0-1 \right)\left( 0-2 \right) \\ & f\left( 0 \right)=0\left( -1 \right)\left( -2 \right) \\ & f\left( 0 \right)=0\,.......................(iii) \\ \end{align}
Now, we will put the values of $f\left( \dfrac{1}{2} \right)\text{ and }f\left( 0 \right)$ from (ii) and (iii) to (i). Thus, we will get:
\begin{align} & \Rightarrow f'\left( c \right)=2\left( \dfrac{3}{8}-0 \right) \\ & \Rightarrow f'\left( c \right)=2\left( \dfrac{3}{8} \right) \\ & \Rightarrow f'\left( c \right)=\dfrac{3}{4}\,........................(iv) \\ \end{align}

Now, we will find $f'\left( x \right)$ i.e. the derivative of f(x). For doing this, we will expand f(x) first. Thus, we will get:
\begin{align} & f\left( x \right)=x\left( x-1 \right)\left( x-2 \right) \\ & \Rightarrow f\left( x \right)=x\left( {{x}^{2}}-x-2x+2 \right) \\ & \Rightarrow f\left( x \right)=x\left( {{x}^{2}}-3x+2 \right) \\ & \Rightarrow f\left( x \right)={{x}^{3}}-3{{x}^{2}}+2x \\ \end{align}
Now $f'\left( x \right)$ will be:
\begin{align} & f'\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{x}^{3}}-3{{x}^{2}}+2x \right] \\ & f'\left( x \right)=3{{x}^{2}}-6x+2 \\ \end{align}

Thus, the value of $f'\left( x \right)$ will be:
$f'\left( c \right)=3{{c}^{2}}-6c+2\,.......................(v)$
From (iv) and (v), we have:
\begin{align} & \Rightarrow 3{{c}^{2}}-6c+2\,=\dfrac{3}{4} \\ & \Rightarrow 3{{c}^{2}}-6c+2-\dfrac{3}{4}\,=0 \\ & \Rightarrow 3{{c}^{2}}-6c+\dfrac{5}{4}\,=0 \\ & \Rightarrow 12{{c}^{2}}-24c+5\,=0 \\ \end{align}
Now, the above equation is a quadratic in c. So, we will find the value of c using the quadratic formula. If the quadratic equation given is $a{{x}^{2}}+bx+c=0$ then we have:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Thus, we have:
\begin{align} & c=\dfrac{-\left( -24 \right)\pm \sqrt{{{\left( -24 \right)}^{2}}-4\left( 12 \right)\left( 5 \right)}}{2\left( 12 \right)} \\ & \Rightarrow c=\dfrac{24\pm \sqrt{576-240}}{24} \\ & \Rightarrow c=\dfrac{24\pm \sqrt{336}}{24} \\ & \Rightarrow c=\dfrac{24\pm 4\sqrt{21}}{24} \\ & \Rightarrow c=\dfrac{6\pm \sqrt{21}}{6} \\ \end{align}
Now, either $c=\dfrac{6+\sqrt{21}}{6}$ or $c=\dfrac{6-\sqrt{21}}{6}$ or both:
If $c=\dfrac{6+\sqrt{21}}{6}$ then $c=\dfrac{6+4.58}{6}$
\begin{align} & \Rightarrow c=\dfrac{10.58}{6} \\ & \Rightarrow c=1.76 \\ \end{align}
If $c=\dfrac{6-\sqrt{21}}{6}$ then $c=\dfrac{6-4.58}{6}$
\begin{align} & \Rightarrow c=\dfrac{1.42}{6} \\ & \Rightarrow c=0.23 \\ \end{align}
So, the value of $c=\dfrac{6-\sqrt{21}}{6}$ because the interval of c is $\left[ 0,\dfrac{1}{2} \right]$

So, the correct answer is “Option C”.

Note: We have not checked whether $f\left( x \right)=x\left( x-1 \right)\left( x-2 \right)$ is continuous or differentiable in $\left[ 0,\dfrac{1}{2} \right]$ This is because f(x) is a cubic polynomial and every cubic polynomial is continuous and differentiable in the $x\in \left( -\alpha ,\alpha \right)$ If we want, we can check this by applying the limit on f(x).