Let `a, n in N` such that `a ge n^(3)`. Then `root3(a+1)-root3(a)` is always

Options

A. less than `(1)/(3n^(2))`
B. less than `(1)/(2n^(3))`
C. more than `(1)/(n^(3))`
D. more than `(1)/(4n^(2))`

Answer

A

Explanation

Let `f(x)=x^(1//3)`
`rArr" "f'(x)=(1)/(3x^(2//3))`
Applying LMVT in `[a,a+1]`, we get one ` cin (a,a+1)`
`f'(c)=(f(a+1)-f(a))/(a+1-a)`
`rArr" "root3(a+1)-root3a=(1)/(3c^(2//3))lt(1)/(3c^(2//3))lt(1)/(3n^(2))`
`rArr" "root3(a+1)-root3alt(1)/(3n^(2))AA a ge n^(3)`