JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
Let $f(x)=x^{3}+x^{2}f'(1)+xf''(2)+f'''(3)$, $x\in\mathbf{R}$. Then $f'(10)$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$202$
Solution
Concept. $f'(1)$, $f''(2)$, $f'''(3)$ are constants; call them $a$, $b$, $c$ and differentiate. Formulas used. (1) $f=x^3+ax^2+bx+c$ $\Rightarrow$ $f'=3x^2+2ax+b$, $f''=6x+2a$, $f'''=6$Step 1 - Equations. By (1), $c=f'''(3)=6$ and $$a=f'(1)=3+2a+b,\qquad b=f''(2)=12+2a.$$ Step 2 - Solve. The first gives $a+b=-3$. With $b=12+2a$: $3a+12=-3$, so $a=-5$, $b=2$. Step 3 - $f'(10)$. $$f'(10)=300+2(-5)(10)+2=202.$$ Answer: 202