यदि `1= sum_(k=1)^(98)int_(k)^(k+1)(k+1)/(x(x+1))dx` तब

Options

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

Explanation

`l=underset(k=1)overset(99)sum int_(k)^(k+1) ((k+1))/(x(x+1)) dx`
स्पष्तः `l gt underset(k=1)overset(98)sum int_(k)^(k+1) ((k+1))/((x+1)^(2)) dx`
`rArr l gt underset(k=1)overset(98)sum (k+1) int_(k)^(k+1) (1)/((x+1)^(2)) dx`
`rArr l gt underset(k=1) overset(98)sum (-k(k+1)) [(1)/(k+2) -(1)/(k+1)]`
`rArr l gt underset(k=1)overset(98)sum (1)/(k+2)`
`rArr l gt 1/3 +.........+1/(100) gt (98)/(100)`
`rArr l gt (49)/(50)`
साथ ही `l lt underset(k=1)overset(98) int_(k)^(k+1) (k+1)/(x(k+1)) dx`
` underset(k=1)overset(98) sum [log_(e) (k+1) -log_(e)k]`
`l lt log_(e) 99`