If f(x) `=underset(x)overset(x^2)int(t-1)dt, 1 le x le 2` then the greatest value of `phi` (x) , is

Options

A. 2
B. 4
C. 8
D. none of these

Answer

B

Explanation

We have ,
`f(x)=underset(x)overset(x^2)int(t-1)dt`
`rArr f(x)=2x(x^2-1)-(x-1)`
`rArr f(x) = 2x^3-3x+1`
`rArr f'(x)=(x-1)(2x^2 +2x -1)`
`rArr f(x)-2(x-1)(x+(sqrt(3)+1)/(2))(x-(sqrt(3)-1)/(2))`
The signs of f(x) for different values for x are shown in Fig.26

Clearly f'(x) `gt0` for all `x in [1,2]` So, f(x) is increasing on [ 1.2]
Hence ,greatest value of f(x) is equal to
`f(2)=underset(2)overset(4)int[(t-1)^2/2]_2^4=4`