Let f(x)= 0x t (t2-9t+20 )dt, 1 x 5. If the range of f is [ , ], then 4( + ) equals :
- 253
- 157
- 154
- 125
Answer
(2) 157
Solution
f'(x)=x(x-4)(x-5) ⇒ f'>0 on (1,4), f'<0 on (4,5) ⇒ maximum at x=4
[figure]
f(x)= x44-3x3+10x2: f(1)= 294, f(4)=32, f(5)= 1254
Min = 294 ⇒ = 294, =32
∴ 4( + )=29+128=157
Answer: (2)