Question

The constant c of Lagrange’s theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is
A. $1\pm \dfrac{2}{\sqrt{3}}$
B. $2\pm \dfrac{2}{\sqrt{3}}$
C. $3\pm \dfrac{2}{\sqrt{3}}$
D. $4\pm \dfrac{2}{\sqrt{3}}$

Answer 1

Hint : According to the Lagrange’s theorem, if a function f(x) is continuous and differentiable over a closed interval [a,b] then, there exist a point ( $x=c$ ) between a and b such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ , where f’(c) is the derivative of the function at $x=c$ and b, a are the boundary values of x.

Complete step-by-step answer :
The given function is $f(x)=(x-1)(x-2)(x-3)$ .
Multiple the brackets in the equation.
$\Rightarrow f(x)=({{x}^{2}}-3x+2)(x-3)$
$\Rightarrow f(x)=({{x}^{3}}-3{{x}^{2}}+2x-3{{x}^{2}}+9x-6)$
$\Rightarrow f(x)={{x}^{3}}-6{{x}^{2}}+11x-6$ …. (i)
This means that the function f(x) is a polynomial function and every polynomial function are continuous and differentiable for all real values of x.
Thus, the given function satisfy the condition for Lagrange’s theorem.
This means that the function f(x) is continuous and differentiable in [0,4].
According to the Lagrange’s theorem, if a function f(x) is continuous and differential over a closed interval [a,b] then, there exist a point ( $x=c$ ) between a and b such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ ,
Where f’(c) is the derivative of the function at $x=c$ and b, a are the boundary values of x.
In this case, $b=4$ and $a=0$ .
$\Rightarrow f'(c)=\dfrac{f(4)-f(0)}{4-0}$ …. (ii)
Substitute $x=a=0$ in (i).
$\Rightarrow f(0)={{0}^{3}}-6{{(0)}^{2}}+11(0)-6=-6$
Now, substitute $x=b=4$ in (i).
$\Rightarrow f(4)={{4}^{3}}-6{{(4)}^{2}}+11(4)-6=64-96+44-6=6$ .
Substitute these values in (ii).
$\Rightarrow f'(c)=\dfrac{6-(-6)}{4}=3$ .
Now, differentiate (i) with respect to x.
$\Rightarrow f'(x)=3{{x}^{2}}-12x+11$
Now put $x=c$ .
$\Rightarrow f'(c)=3{{c}^{2}}-12c+11$ .
But, $f'(c)=3$ .
$\Rightarrow 3{{c}^{2}}-12c+11=3$
$\Rightarrow 3{{c}^{2}}-12c+8=0$ .
By using quadratic formula we get that
$\Rightarrow c=\dfrac{-(-12)\pm \sqrt{{{(-12)}^{2}}-4(3)(8)}}{2(3)}$
$\Rightarrow c=\dfrac{12\pm \sqrt{144-96}}{6}=\dfrac{12\pm \sqrt{48}}{6}$
$\Rightarrow c=\dfrac{12\pm 4\sqrt{3}}{6}=2\pm \dfrac{2\sqrt{3}}{3}=2\pm \dfrac{2}{\sqrt{3}}$ .
So, the correct answer is “Option B”.

Note : For the Lagrange’s theorem to be valid for a given function, the function must be continuous and differentiable over the mentioned closed interval. It must be continuous as well as differentiable at the boundary values of x (a and b).