Let $\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4]$, Values of C will be _______ [if L.M.V.T. (Lagrange's Mean…

Let $\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4]$, Values of C will be _______ [if L.M.V.T. (Lagrange's Mean Value Theorem) can be applied].
  1. $\frac{4-2 \sqrt{3}}{3}, \frac{4+2 \sqrt{3}}{3}$
  2. $\frac{6-2 \sqrt{3}}{3}, \frac{6+2 \sqrt{3}}{3}$
  3. $\frac{6-\sqrt{3}}{3}, \frac{6+\sqrt{3}}{3}$
  4. $2-\sqrt{3}, 2+\sqrt{3}$

Solution

$\begin{aligned} & \text {Let } y=(x-1)(x-2)(x-3) \\ & \log y=\log (x-1)+\log (x-2)+\log (x-3) \end{aligned}$
Differentiating w.r.t. $x$, we get $\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}$ $\begin{aligned} & \therefore \quad \mathrm{f}^{\prime}(x)=\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-2)(x-3)+(x-1)(x-3) \\ &+(x-1)(x-2) \end{aligned}$ $\begin{array}{ll}\therefore & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(4)-\mathrm{f}(0)}{4-0}=\frac{6-(-6)}{4}=3, \text { for } \mathrm{c} \in[0,4] \\ \therefore \quad & (\mathrm{c}-2)(\mathrm{c}-3)+(\mathrm{c}-1)(\mathrm{c}-3)+(\mathrm{c}-1)(\mathrm{c}-2)=3 \\ \therefore \quad & 3 \mathrm{c}^2-12 \mathrm{c}+11=3 \\ \therefore \quad & 3 \mathrm{c}^2-12 \mathrm{c}+8=0 \\ \therefore \quad & \mathrm{c}=\frac{12 \pm \sqrt{144-96}}{6}=\frac{6 \pm 2 \sqrt{3}}{3}\end{array}$

Asked in: MHT CET 2024 (10 May Shift 2)

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