If L.M.V. theorem is true for f(x) = x(x-1)(x-2);, x ∈ [0,, 1/2] , then C = ?

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5. Continuity and Differentiation

hard

If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?

A

$\left( {1 + \frac{{\sqrt {21} }}{6}} \right)$

B

$\left( {1 - \frac{{\sqrt {21} }}{6}} \right)$

C

$\frac {3}{16}$

D

Doe not exist

Solution

$f(x)=x(x-1)(x-2)=x\left(x^{2}-3 x+2\right)$

$f(x)=x^{3}-3 x^{2}+2 x$

$LMVT,$

$3 x^{2}-6 x+2=\frac{f(1 / 2)-f(0)}{1 / 2}$

$3{x^2} – 6x + 2 = \frac{{\frac{1}{2} \times – \frac{1}{2} \times – \frac{3}{2} – 0}}{{1/2}}$

$3 x^{2}-6 x+2=\frac{3}{4}$

$12 x^{2}-24 x+5=0$

$x=\frac{24 \pm \sqrt{144 \times 4-4 \times 12 \times 5}}{2 \times 12}$

$x=\frac{24 \pm 2 \sqrt{84}}{24}$

$x=1 \pm \frac{4 \sqrt{21}}{24}$

$x=1-\frac{\sqrt{21}}{6}$

Standard 12

Mathematics

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