The value of c in Lagrange's mean value theorem for the function … - Vidyadip

MCQ

The value of $c$ in Lagrange's mean value theorem for the function $f(x) = x(x - 2)$ when $\text{x}\in[1,2]$ is :

A
$1$
B
$\frac{1}{2}$
C
$\frac{2}{3}$
✓
$\frac{3}{2}$

✓

Answer

Correct option: D.

$\frac{3}{2}$

We have
$f(x) = x(x - 2)$
It can be rewritten as $f(x) = x^2 - 2x$
We know that a polynomial function is everywhere continuous and differentiable.
Since, $f(x)$ is polynomial, it is continuous on $[1, 2]$ and differentiable on $[1, 2].$
So, there must exist at least one real number $\text{c}\in(1,2)$ such that
$\text{f}\ '(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$=\frac{\text{f}(2)-\text{f}(1)}{1}$
Now, $f(x) = x^2 - 2x$
$\Rightarrow f'(x) = 2x - 2$
and $f(1) = -1, f(2) = 0$
$\therefore\ \text{f}\ '(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow\text{f}(\text{x})=\frac{0+1}{1}$
$\Rightarrow2\text{x}-2=1$
$\Rightarrow\text{x}=\frac{3}{2}$
$\therefore\ \text{c}=\frac{3}{2}\in(1,2)$

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