JEE Main 2026 (Online) 21st January Evening ShiftQuestion 1
Let $f(x)=x^3+x^2f'(1)+2xf''(2)+f'''(3)$, $x\in\mathbf{R}$. Then the value of $f'(5)$ is :
Differentiation and its Application
Solution
Concept. $f'(1)$, $f''(2)$, $f'''(3)$ are constants; call them $A$, $B$, $C$ and differentiate. Formulas used. (1) $f(x)=x^3+Ax^2+2Bx+C$ $\Rightarrow$ $f'=3x^2+2Ax+2B$, $f''=6x+2A$, $f'''=6$Step 1 - Equations. By (1): $$A=f'(1)=3+2A+2B,\qquad B=f''(2)=12+2A.$$ Step 2 - Solve. The first gives $A+2B=-3$. With $B=12+2A$: $5A+24=-3$, so $A=-\frac{27}{5}$ and $B=12-\frac{54}{5}=\frac65$. Step 3 - $f'(5)$. $$f'(5)=75+10A+2B=75-54+\frac{12}{5}=\frac{117}{5}.$$ Answer: (A)