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

Options

A. `1- sqrt(15)/(6)`
B. `1+ sqrt(15)`
C. `1 - sqrt(21)/(6)`
D. ` 1+ sqrt(21)`

Answer

C

Explanation

Form mean value therorem,
`f'(c) = (f(b) -f(a))/(b-a)`
Given, `a = 0 rArr f(a) = 0`
`b = (1)/(2) rArr f(b) = (3)/(8)`
Now `f'(x) = (x-1)(x+2)+x(x-2)+x(x-1)`
`f'(c) = (c-1)(c-2)+c(c-2)+c(c-1)`
` = c^(2) -3c +2 + c^(2) - 2c + c^(2) - c`
`rArr f'(c) = 3(c)^(2) - 6c + 2`
By dfinition of mean vlaue of theorm `f(C) = (f(b) -f(a))/(b-a)`
`rArr 3c^(2) - 6c + 2 =((3)/(8) -0)/((1)/(2) -0) = (3)/(4)`
`rArr 3c^(2) - 6c + (5)/(4) = 0`
The is a quadratic equation in c.
`therefore " " c = (6pm sqrt(36 - 15))/(2 xx3)`
` = (6 pm sqrt(21))/(6) = 1 pm (sqrt(21))/(6)`
Since 'c' lies in the interval `[0,(1)/(2)]`
`therefore" " c = 1 - (sqrt(21))/(6)" "["neglecting"c = 1 + sqrt(21)/(6)\]`