The equation of the curve is `y=f(x)dot` The tangents at `[1,f(1),[2,f(2)],a n d[3,f(3)]` make angles `pi/6,pi/3,a n dpi/4,` respectively, with the positive direction of x-axis. Then the value of `int_2^3f'(x)f''(x)dx+int_1^3f''(x)dx` is equal to

Options

A. `-1//sqrt(3)`
B. `1//sqrt(3)`
C. `0`
D. none of these

Answer

A

Explanation

Given `f'(1)=tan pi//6, f'(2)=tanpi//3, f'(3)=tan pi//4`
Now `int_(2)^(3)f'(x)f''(x)dx+int_(1)(3)f''(x)dx`
`=[((f'(x))^(2))/2]_(2)^(3)+[f'(x)]_(1)^(3)`
`=((f'(3))^(2)-(f(2))^(2))/2+f'(3)-f'(1)`
`=((1)^(2)-(sqrt(3))^(2))/2+(1-1/(sqrt(3)))`
`=(1-3)/2+1-1/(sqrt(3))`
`=(-1)/(sqrt(3))`