Let f(x) = x3 + x2 f'(1) + 2x f''(2) + f'''(3), x R. Then the value of f'(5) is :
- 6575
- 1175
- 25
- 625
Answer
(2) 1175
Solution
f'(1), f''(2) and f'''(3) are constants. Let f'(1) = a, f''(2) = b, f'''(3) = c.
f(x) = x3 + ax2 + 2bx + c
f'(x) = 3x2 + 2ax + 2b, f''(x) = 6x + 2a, f'''(x) = 6
Put x = 1 in f'(x): a = 3 + 2a + 2b a + 2b = -3 ...(i)
Put x = 2 in f''(x): b = 12 + 2a ...(ii)
From (i) and (ii): a + 24 + 4a = -3
a = - 275, b = 12 - 545 = 65
f'(5) = 75 + 10a + 2b
= 75 - 54 + 125
= 1175
Answer: (2)