Let f(x) be a continuously differentiable function on the interval (0, ) such that f(1)=2 and t x t10f(x)-x10f(t)t9-x9=1 for each x>0. Then, for all x>0, f(x) is equal to
- 3111x- 911x10
- 911x+ 1311x10
- -911x+ 3111x10
- 1311x+ 911x10
Answer
(2) 911x+ 1311x10
Solution
The limit is of the form 00; apply L'Hospital's rule w.r.t. t.
t x 10t9f(x)-x10f'(t)9t8= 10x9f(x)-x10f'(x)9x8=1
10xf(x)-x2f'(x)=9 f'(x)- 10xf(x)=- 9x2
I.F. =x-10: ddx (x-10f )=-9x-12 x-10f= 911x-11+C
f(x)= 911x+Cx10; f(1)=2 C= 1311
Answer: (B)