JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
If Rolle's theorem holds for the function $f(x)=x^{3}-ax^{2}+bx-4$, $x\in[1,2]$ with $f'\left(\frac{4}{3}\right)=0$, then ordered pair $(a,b)$ is equal to :
Differentiation and its Application
Solution
Concept. Rolle's theorem needs $f(1)=f(2)$; the given $f'\left(\frac43\right)=0$ is the second equation. Formulas used. (1) $f'(x)=3x^2-2ax+b$Step 1 - $f(1)=f(2)$. $-3-a+b=4-4a+2b$, so $3a-b=7$. Step 2 - $f'\left(\frac43\right)=0$. By (1), $\frac{16}{3}-\frac{8a}{3}+b=0$, i.e. $16-8a+3b=0$. Step 3 - Solve. $b=3a-7$: $16-8a+9a-21=0$, so $a=5$, $b=8$. Answer: (A)