JEE Main 2019 (Online) 10th January Morning ShiftQuestion 1
Let $f:R\to R$ be a function such that $f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3),\ \forall x\in R$. Then $f(2)$ equals
Differentiation and its Application
Solution
Concept. $f'(1)$, $f''(2)$, $f'''(3)$ are just constants; call them $a,b,c$ and match. Formulas used. (1) For $f=x^3+ax^2+bx+c$: $f'=3x^2+2ax+b$, $f''=6x+2a$, $f'''=6$Step 1 - $c$. $c=f'''(3)=6$. Step 2 - $a$ and $b$. By (1): $$a=f'(1)=3+2a+b,\qquad b=f''(2)=12+2a.$$ Substituting: $a=15+4a$, so $a=-5$ and $b=2$. Step 3 - $f(2)$. $8-20+4+6=-2$. Answer: (D)