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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems4 Marks
Question
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
f(x) = x(x - 1) on [1, 2]
✓

Answer

We have
f(x) = x(x - 1) which can be rewritten as f(x) = x2 - x
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [1, 2] and differentiable on (1, 2).
Thus, both conditions of Lagrange's mean value theorem is satisfied.
So, there must exist at least one real number $\text{c}\in(1,2)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
Now, f(x) = x2 - x
⇒ f'(x) = 2x - 1,
⇒ f(2) = 2, f(1) = 0
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow2\text{x}-1=\frac{2-0}{2-1}$
$\Rightarrow2\text{x}-1-2=0$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Thus, $\text{c}=\frac{3}{2}\in(1,2)$ such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
Hence, Lagrange's mean value theorem is verified.

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