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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify mean value theorem for the function:
$\text{f(x)}=\text{x}^3-2\text{x}^2-\text{x}+3\text{ in }[0,1].$
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Answer

Consider, $\text{f(x)}=\text{x}^3-2\text{x}^2-\text{x}+3\text{ in }[0,1]$
Since, f(x) is a polynomial function
Hence, f(x) is continuous in [0,1]
$\text{f(x)}=3\text{x}^2-4\text{x}-1$ which ecists in (0, 1).
Hence, f(x) is differentiable in (0, 1).
Since, conditions of mean value theorem are satisfied.
Therefore, by mean value theorem $\exists\text{ c}\in(0,1),$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\ 3\text{c}^2-4\text{c}-1=\frac{[1-2-1+3]-[0+3]}{1-0}$
$\Rightarrow\ 3\text{c}^2-4\text{c}-1=\frac{-2}{1}$
$\Rightarrow\ 3\text{c}^2-4\text{c}+1=0$
$\Rightarrow\ 3\text{c}^2-3\text{c}-\text{c}+1=0$
$\Rightarrow\ 3\text{c}(\text{c}-1)-1(\text{c}-1)=0$
$\Rightarrow\ (3\text{c}-1)(\text{c}-1)=0$
$\Rightarrow\ \text{c}=\frac{1}{3},1,$ where $\frac{1}{3}\in(0,1)$
Hence, the mean value theorem has been verified.

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