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Question: In the mean-value theorem \(\frac{f(b) - f(a)}{b - a} = f^{'}(c)\), if \(a = 0\), \(b = \frac{1}{2}\...

In the mean-value theorem f(b)−f(a)b−a=f′(c)\frac{f(b) - f(a)}{b - a} = f^{'}(c), if a=0a = 0, b=12b = \frac{1}{2} and f(x)=x(x−1)(x−2),f(x) = x(x - 1)(x - 2), the value of c is

A

1−1561 - \frac{\sqrt{15}}{6}

B

1+151 + \sqrt{15}

C

1−2161 - \frac{\sqrt{21}}{6}

D

1+211 + \sqrt{21}

Answer

1−2161 - \frac{\sqrt{21}}{6}

Explanation

Solution

From mean value theorem f′(c)=f(b)−f(a)b−af^{'}(c) = \frac{f(b) - f(a)}{b - a}

a=0,f(a)=0a = 0,f(a) = 0 ⇒ b=12,f(b)=38b = \frac{1}{2},f(b) = \frac{3}{8}

f′(x)=(x−1)(x−2)+x(x−2)+x(x−1)f^{'}(x) = (x - 1)(x - 2) + x(x - 2) + x(x - 1),

f′(c)=(c−1)(c−2)+c(c−2)+c(c−1)f^{'}(c) = (c - 1)(c - 2) + c(c - 2) + c(c - 1)

= c2−3c+2+c2−2c+c2−cc^{2} - 3c + 2 + c^{2} - 2c + c^{2} - c, f′(c)=3c2−6c+2f^{'}(c) = 3c^{2} - 6c + 2

According to mean value theorem

⇒ f′(c)=f(b)−f(a)b−af^{'}(c) = \frac{f(b) - f(a)}{b - a}

⇒ 3c2−6c+2=(38)−0(12)−0=34⇒3c2−6c+54=03c^{2} - 6c + 2 = \frac{\left( \frac{3}{8} \right) - 0}{\left( \frac{1}{2} \right) - 0} = \frac{3}{4} \Rightarrow 3c^{2} - 6c + \frac{5}{4} = 0

c=6±36−152×3=6±216=1±216c = \frac{6 \pm \sqrt{36 - 15}}{2 \times 3} = \frac{6 \pm \sqrt{21}}{6} = 1 \pm \frac{\sqrt{21}}{6}.