Let f(x)=x3+x2f'(1)+2xf''(2)+f'''(3), x∈R. Then the value of f'(5) is [2026]
(1)
f'(x)=3x2+2xf'(1)+2f''(2)
f''(x)=6x+2f'(1)
f''(2)=12+2f'(1)
∴ f'(x)=3x2+2xf'(1)+2(12+2f'(1))
f'(x)=3x2+2(x+2)f'(1)+24
Putting x=1
f'(1)=3+6f'(1)+24
-5f'(1)=27⇒f'(1)=-275
∴ f''(2)=12+2(-275)=12-545=65
∴ f'(x)=3x2-545x+125
∴ f'(5)=75-54+125=1175