Q128 / 200 · Manipal (MET) 2014 · Mathematics
What are the values of $c$ for which Rolle's theorem for the function $f(x)=x^3-3 x^2+2 x$ in the interval $[0,2]$ is verified?
- A$c= \pm 1$
- Correct answerB$c=1 \pm \frac{1}{\sqrt{3}}$
- C$c= \pm 2$
- DNone of these
Solution
Here, we observe that
(a) $f(x)$ is a polynomial, so it is continuous in the interval $[0,2]$.
(b) $f^{\prime}(x)=3 x^2-6 x+2$ exists for all $x \in(0,2)$.
So, $f(x)$ is differentiable for all $x \in(0,2)$ and (c) $f(0)=0, f(2)=2^3-3(2)^2+2(2)=0$ $\Rightarrow f(0)=f(2)$
Thus, all the three conditions of Rolle's theorem are satisfied.
So, there must exist $c \in[0,2]$ such that $f^{\prime}(c)=0$ $\begin{array}{ll} \Rightarrow & f^{\prime}(c)=3 c^2-6 c+2=0 \\ \Rightarrow & c=1 \pm \frac{1}{\sqrt{3}} \in[0,2] \end{array}$
So, $f(x)$ is differentiable for all $x \in(0,2)$ and (c) $f(0)=0, f(2)=2^3-3(2)^2+2(2)=0$ $\Rightarrow f(0)=f(2)$
Thus, all the three conditions of Rolle's theorem are satisfied.
So, there must exist $c \in[0,2]$ such that $f^{\prime}(c)=0$ $\begin{array}{ll} \Rightarrow & f^{\prime}(c)=3 c^2-6 c+2=0 \\ \Rightarrow & c=1 \pm \frac{1}{\sqrt{3}} \in[0,2] \end{array}$
Hence, the correct answer is option: 'B'.