Question

The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is
(a) 1
(b) 1/2
(c) 2/3
(d) 3/2

Answer 1

(d)$\frac{3}{2}$

We have
f (x) = x (x − 2)

It can be rewritten as $f\left(x\right)=x^2-2x$.

We know that a polynomial function is everywhere continuous and differentiable.

Since $f\left(x\right)$ is a polynomial , it is continuous on $\left[1,2\right]$ and differentiable on $\left(1,2\right)$.

Thus, $f\left(x\right)$ satisfies both the conditions of Lagrange's theorem on $\left[1,2\right]$.

So, there must exist at least one real number c $\in \left(1,2\right)$ such that

$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{f\left(2\right)-f\left(1\right)}{1}$

Now, $f\left(x\right)=x^2-2x$
$\Rightarrow f\text{'}\left(x\right)=2x-2$,
and $f\left(1\right)=-1,f\left(2\right)=0$

$\therefore f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$
$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{0+1}{1} \\ \Rightarrow 2x-2=1 \\ \Rightarrow x=\frac{3}{2} \end{gathered}$$

∴ $c=\frac{3}{2}\in \left(1,2\right)$