Question

For the function $f(x)=(x-1)(x-2)$ defined on $\left[0,\frac{1}{2}\right]$, the value of $c$ satisfying Lagrange's mean value theorem is

• A. $\frac{1}{5}$
• B. $\frac{1}{3}$
• C. $\frac{1}{7}$
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

We have a function
$f(x)=(x-1)(x-2)$
$=x^2-3x+2$ defined on $\left[0,\frac{1}{2}\right]$
According to LMV theorem,
$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{1}{2}\right)-f(0)}{\frac{1}{2}-0}$
$\Rightarrow 2c-3=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)-(-1)(-2)}{\frac{1}{2}}$
$\Rightarrow 2c-3=2\left(\frac{3}{4}-2\right)$
$\Rightarrow 2c-3=-\frac{5}{2}$
$\Rightarrow 4c-6=-5$
$\Rightarrow c=\frac{1}{4}\in \left(0,\frac{1}{2}\right)$
Thus, required value of $c$ is $\frac{1}{4}$.