If `I=sum_(k=1)^(98)int_(k)^(k+1)(k+1)/(x(x+1))dx`, then

Options

A. `I gt log_(e )99`
B. `I lt log_(e )99`
C. `I lt (49)/(50)`
D. `I gt(49)/(50)`

Answer

B, D

Explanation

For any `x in [k,k+1],k inN`, we know that
`(1)/(x+1) lt (k(k+1))/(x(x+1)) lt (1)/(x)`
`rArr overset(k+1)underset(k)int (1)/(x+1)dx lt overset(k+1)underset(k)int (k(k+1))/(x(x+1))dx lt overset(k+1)underset(k)int (1)/(x)dx`
`rArr overset(98)underset(k=1)sumlog_(e )overset(k+1)underset(k)int (1)/(x+1)dx lt overset(98)underset(k=1)sum overset(k+1)underset(k)int(k(k+1))/(x(x+1))dx lt overset(98)underset(k=1)sum overset(k+1)underset(k)int(1)/(x)dx`
`rArr overset(98)underset(k=1)sum{log_(e)(k+2)-log_(e )(k+1)} lt I lt overset(98)underset(k=1)sum{log_(e )(k+1)-log_(e )k}`
`rArr log_(e )100-log_(e )2 lt I lt log_(e )99-log_(e )1`
`rArr log_(e )50 lt I lt log_(e )99`