If `Isum_(k=1)^(98)int_k^(k+1)(k+1)/(x(x+1))dx ,t h e n:`

Options

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

Answer

B, D

Explanation

We have `sum_(k=1)^(98)int_(k)^(k+1)1/(x+1)dxlt sum_(k=1)^(98)int_(k)^(k+1)(k+1)/(x(x+1))dx lt sum_(k=1)^(98)int_(k)^(k+1)(dx)/x`
`implies sum_(k=1)^(98)(log_(e)(k+2)-log_(e)(k+1))ltI`
`ltsum_(k=1)^(98)(log_(e)(k+1)-log_(e)k)`
`implieslog_(e)50ltI I lt log_(e)99`
`implies49/50lt I lt log_(e)99`
Alternative method:
Put `x-k=p`
`implies I=sum_(k=1)^(98)int_(0)^(1)(k+1)((k+p)(k+p+1))dp`
Now `Igtsum_(k=1)^(98)int_(0)^(1)(k+1)/((k+p+1)^(2))dp`
`implies Igt sum_(k=1)^(98)(k+1)((-1)/(k+p+q))_(0)^(1)`
`implies Igtsum_(k=1)^(98)(k+1)(1/(k+1)-1/(k+2))`
`implies Igtsum_(k=1)^(98)1/(k+2)=1/3+......+1/100`
`impliesIgt1/100+.......+1/100=98/100`
`impliesIgt49/50`
Now `(k+1)/(x(x+1))lt(k+1)/(x(k+1)) ( :' "least value of" x+1 "is"k+1)`
`implies(k+1)/(x(x+1)lt1/x)`
`implies Ilt lsum_(k=1)^(98)int_(k)^(k+1)1/x dx`
`implies Iltsum_(k=1)^(98)log_(e)(kK+1)-log_(e)k`
`implies1ltlog_(e)99`