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Mean Value Theorems — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMean Value Theorems5 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 $f(x) = x^2 - x$
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, $f(x) = x^2 - x$
$\Rightarrow f'(x) = 2x - 1,$
$\Rightarrow f(2) = 2, f(1) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow2\text{x}-1=\frac{2-0}{2-1}$
$\Rightarrow2\text{x}-1-2=0$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
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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