Let f(x)=x3+x2f'(1)+xf''(2)+f'''(3), x∈R. Then f'(10) is equal to _____ . [2024]
(202)
f(x)=x3+x2f'(1)+xf''(2)+f'''(3), x∈R
f'(x)=3x2+2xf'(1)+f''(2),
f''(x)= 6x+2f'(1)
f'''(x)=6⇒f'''(3)=6
f''(2)=6(2)+2f'(1)=12+2f'(1),
f'(1)=3(1)2+2(1)f'(1)+f''(2)=3+2f'(1)+f''(2)
⇒f'(1)=3+2f'(1)+12+2f'(1)
⇒3f'(1)=-15⇒f'(1)=-5
∴ f''(2)=12+2(-5)=2
Now, f'(10)=3(10)2+2(10)f'(1)+f''(2)
=300+20(-5)+2=202