JEE Main 2020 (Online) 7th January Evening ShiftQuestion 1
The value of $c$, in the Lagrange's mean value theorem for the function $f(x)=x^3-4x^2+8x+11$, when $x\in[0,1]$, is :
Differentiation and its Application
Solution
Concept. Solve $f'(c)=\frac{f(1)-f(0)}{1-0}$ and keep the root in $(0,1)$. Formulas used. (1) $f'(x)=3x^2-8x+8$Step 1 - Chord slope. $f(1)-f(0)=16-11=5$. Step 2 - Solve. By (1), $3c^2-8c+8=5$, i.e. $3c^2-8c+3=0$, so $c=\frac{4\pm\sqrt7}{3}$. Step 3 - Choose. $\frac{4+\sqrt7}{3}>1$, so $c=\frac{4-\sqrt7}{3}$. Answer: (B)