Let f be a real valued continuous function defined on the positive real axis such that g(x) = 0x t f(t) dt. If g (x3 ) = x6 + x7, then value of r=115 f (r3 ) is :
- 270
- 340
- 310
- 320
Answer
(3) 310
Solution
g(x) = 0x t f(t) dt g'(x) = x f(x)
Differentiating g(x3) = x6 + x7:
g'(x3) 3x2 = 6x5 + 7x6
x3 f(x3) 3x2 = 6x5 + 7x6
f(x3) = 6x5 + 7x63x5 = 2 + 7x3
r=115 f(r3) = r=115 (2 + 7r3 )
= 30 + 73 15 162
= 30 + 73 120 = 30 + 280 = 310
Answer: (3)