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5. Continuity and Differentiation
hard
The value of $c$ in the Lagrange's mean value theorem for the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11$ when $\mathrm{x} \in[0,1]$ is
A
$\frac{2}{3}$
B
$\frac{\sqrt{7}-2}{3}$
C
$\frac{4-\sqrt{5}}{3}$
D
$\frac{4-\sqrt{7}}{3}$
(JEE MAIN-2020)
Solution
$f(0)=11$
$f(1)=16$
$\frac{\mathrm{f}(1)-\mathrm{f}(0)}{1-0}=3 \mathrm{c}^{2}-8 \mathrm{c}+8$
$\Rightarrow 3 c^{2}-8 c+8=5$
$\Rightarrow 3 c^{2}-8 c+3=0$
$c \in[0,1] \Rightarrow c=\frac{4-\sqrt{7}}{3}$
Standard 12
Mathematics
