The value of c for which Rolle's theorem for the function $\mathrm{f}(x)=x^3-3 x^2+2 x$ in the interval $[0…
The value of c for which Rolle's theorem for the function $\mathrm{f}(x)=x^3-3 x^2+2 x$ in the interval $[0,2]$ are
- $\pm 1$
- $\pm 2$
- $1 \pm \frac{1}{\sqrt{3}}$
- $\sqrt{3}(1 \pm \sqrt{3})$
Solution
$\begin{aligned} & \mathrm{f}(x)=x^3-3 x^2+2 x \\ & \mathrm{f}^{\prime}(x)=3 x^2-6 x+2 \\ & \text { Now, } \mathrm{f}^{\prime}(\mathrm{c})=0 \\ & \Rightarrow 3 \mathrm{c}^2-6 \mathrm{c}+2=0 \\ & \Rightarrow \mathrm{c}=\frac{6 \pm \sqrt{12}}{6} \\ & \Rightarrow \mathrm{c}=1 \pm \frac{\sqrt{12}}{6} \\ & \Rightarrow \mathrm{c}=1 \pm \frac{1}{\sqrt{3}}\end{aligned}$
Asked in: MHT CET 2024 (03 May Shift 2)
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