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), the value of c is -

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), the value of c is - (A) 1-(√15/6) (B) 1+√15 (C) 1-(√21/6) (D) 1+√21

Tardigrade Admin · Accepted Answer

f '(c)=(f (b)-f (a)/b-a) ⇒ 3c2-6c+2=(3/8-0/1/2-0)=(3/4) ⇒ c=1±(√21/6) ⇒ c=1+(√21/6) ∉ (0, (1/2)) ⇒ c=1-(√21/6)

Q. In the mean value theorem $\frac{f ( b ) - f ( a )}{b - a} = f^{'} (c),$ if $a = 0, b = 1/2$ and $f (x) = x (x - 1) (x - 2)$ , the value of $c$ is -

$1 - \frac{15}{6}$

$1 + 15$

$1 - \frac{21}{6}$

$1 + 21$

$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$
$\Rightarrow 3 c^{2} - 6 c + 2 = \frac{3/8 - 0}{1/2 - 0} = \frac{3}{4}$
$\Rightarrow c = 1 \pm \frac{21}{6}$
$\Rightarrow c = 1 + \frac{21}{6} \in / (0, \frac{1}{2})$
$\Rightarrow c = 1 - \frac{21}{6}$

Q. In the mean value theorem b−af(b)−f(a)​=f′(c), if a=0,b=1/2 and f(x)=x(x−1)(x−2), the value of c is -

1−615​​

1+15​

1−621​​

1+21​