Q.

Let f(x)=x3+x2f'(1)+2xf''(2)+f'''(3),  xR. Then the value of f'(5) is             [2026]

1 1175  
2 25  
3 625  
4 6575  

Ans.

(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)=27f'(1)=-275

 f''(2)=12+2(-275)=12-545=65

 f'(x)=3x2-545x+125

 f'(5)=75-54+125=1175