JEE PYQ: Application Of Derivatives Question 45
Question 45 - 2020 (07 Jan Shift 2)
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:
(1) $\frac{4 - \sqrt{5}}{3}$
(2) $\frac{4 - \sqrt{7}}{3}$
(3) $\frac{2}{3}$
(4) $\frac{\sqrt{7} - 2}{3}$
Show Answer
Answer: (2)
Solution
$f’(c) = \frac{f(1) - f(0)}{1} = 16 - 11 = 5$. $3c^2 - 8c + 8 = 5 \Rightarrow 3c^2 - 8c + 3 = 0 \Rightarrow c = \frac{8 \pm \sqrt{64-36}}{6} = \frac{4 \pm \sqrt{7}}{3}$. Taking $c \in [0,1]$: $c = \frac{4-\sqrt{7}}{3}$.
BETA
