If Mean value theorem holds for the function $\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4] \quad$ then the…
If Mean value theorem holds for the function $\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4] \quad$ then the values of c as per the theorem are
- $2 \pm \frac{4}{\sqrt{3}}$
- $2 \pm \frac{2}{\sqrt{3}}$
- $2 \pm \sqrt{2}$
- $2 \pm \sqrt{3}$
Solution
$\begin{aligned} & \mathrm{f}(x)=(x-1)(x-2)(x-3) \\ & \mathrm{f}(4)=(4-1)(4-2)(4-3) \\ & \mathrm{f}(4)=6 \\ & \mathrm{f}(0)=(0-1)(0-2)(0-3)=-6\end{aligned}$
$\therefore \quad$ Using LMVT
$f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}=\frac{6-(-6)}{4}=3$
$\begin{aligned}
& \therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=3 ...(i)\\
& \text { Now, } \mathrm{f}(x)=(x-1)(x-2)(x-3) \\
&=x^3-6 x^2+11 x-6
\end{aligned}$
$\begin{aligned}
\therefore \quad & f^{\prime}(x)=3 x^2-12 x+11 \\
& f^{\prime}(\mathrm{c})=3 \\
\Rightarrow & 3 \mathrm{c}^2-12 \mathrm{c}+11=3 \\
\Rightarrow & 3 \mathrm{c}^2-12 \mathrm{c}+8=0 \\
\Rightarrow & \mathrm{c}=\frac{12 \pm \sqrt{48}}{6}=2 \pm \frac{2 \sqrt{3}}{3}=2 \pm \frac{2}{\sqrt{3}}
\end{aligned}$
...[From (i)]
Asked in: MHT CET 2024 (09 May Shift 1)
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