Question

in the mean value theorem${ f(b) -f(a) }/b-a$=f'c if a=0, b=1/2 and f(x) =x(x-1) (x-2) , find the value of c?
Help me​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = x(x - 1)(x - 2) \: \: on \: \: [0, \: \dfrac{1}{2} \bigg]$

Step :- 1

$\rm :\longmapsto\:f(x) \: is \: polynomial \:$

$\bf\implies\:f(x) \: is \:continuous\: on \: [0, \dfrac{1}{2} \bigg]$

Step :- 2

$\rm :\longmapsto\:f(x) = x(x - 1)(x - 2)$

$\rm :\longmapsto\:f(x) = x( {x}^{2} - 2x - x + 2)$

$\rm :\longmapsto\:f(x) = x( {x}^{2} - 3x+ 2)$

$\rm :\longmapsto\:f(x) = {x}^{3} - 3 {x}^{2} + 2x$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:f'(x) = 3{x}^{2} - 6{x} + 2$

$\rm :\longmapsto\:f'(x) \: is \: polynomial \:$

$\bf\implies\:f(x) \: is \:differentiable\: on \: (0, \dfrac{1}{2} \bigg)$

Hence, Mean Value Theorem is applicable.

Now,

$\rm :\longmapsto\:f(x) = x(x - 1)(x - 2) \:$

So,

$\rm :\longmapsto\:f(0) = 0$

and

$\rm :\longmapsto\:f(\dfrac{1}{2} ) = \dfrac{1}{2} (\dfrac{1}{2} - 1)(\dfrac{1}{2} - 2) \:$

$\rm :\longmapsto\:f(\dfrac{1}{2} ) = \dfrac{1}{2} ( - \dfrac{1}{2})( - \dfrac{3}{2}) \:$

$\rm :\longmapsto\:f(\dfrac{1}{2} ) = \dfrac{3}{8}$

Now, Mean Value Theorem is applicable. So, there exist atleast one real number c belongs to (0, 1/2) such that

$\rm :\longmapsto\:f'(c) = \dfrac{f(b) - f(a)}{b - a}$

$\rm :\longmapsto\: {3c}^{2} - 6c + 2 = \dfrac{\dfrac{3}{8} - 0}{\dfrac{1}{2} - 0}$

$\rm :\longmapsto\: {3c}^{2} - 6c + 2 = \dfrac{3}{4}$

$\rm :\longmapsto\: {12c}^{2} - 24c + 8 = 3$

$\rm :\longmapsto\: {12c}^{2} - 24c +5 = 0$

Now, to get the value of 'c', we have to use Quadratic Formula,

$\rm :\longmapsto\:c = \dfrac{ - ( - 24) \: \pm \: \sqrt{ {( - 24)}^{2} - 4 \times 12 \times 5} }{2 \times 12}$

$\rm :\longmapsto\:c = \dfrac{24\: \pm \: \sqrt{576 - 240 }}{24}$

$\rm :\longmapsto\:c = \dfrac{24\: \pm \: \sqrt{336 }}{24}$

$\rm :\longmapsto\:c = \dfrac{24\: \pm \: 4\sqrt{21 }}{24}$

$\rm :\longmapsto\:c = 1 \: \pm \: \dfrac{ \: \sqrt{21 }}{6}$

$\rm :\longmapsto\:c = 1 \: - \: \dfrac{ \: \sqrt{21 }}{6} \: as \: c \in \: (0, \dfrac{1}{2} \bigg)$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (c) \: is \: correct}}}$