a) Given f(x) = ln(x), use the Mean Value Theorem to show that for 0 <a <b, (b-a)/b < ln(b/a) < (b-a)/a. b)Hence show that ln(1.2) lies between 1/m and 1/n, with m, n consecutive integers to be determined.

a) f(x)=ln(x), so f'(x)=1/x. By MVT, f'(c) = (f(b)-f(a))/(b-a) = (ln b - ln a)/(b-a) = ln(b/a)/(b-a), where c lies between a and b.  Now, since 1/x is a decreasing function, and a < c < b, we get 1/b < 1/c < 1/a; i.e. f'(b)

b) Using the inequality above and letting b=1.2, a=1, gives:

(1.2-1)/1.2 < ln(1.2) < 1.2-1, so 0.2/1.2 < ln(1.2) < 0.2;  1/6 < ln(1.2) <1/5, thus m=6, n=5.