Question

If the mean value theorem is $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$. Then, for the function $x^2-2x+3$ in $\left[1,\frac{3}{2}\right]$, the value of $c$ is.

• A. $\frac{6}{5}$
• B. $\frac{5}{4}$
• C. $\frac{4}{3}$
• D. $\frac{7}{6}$

Answer 1

The correct option is B

$\frac{5}{4}$

Explanation of the correct option.

Compute the required value.

Given : $f\left(x\right)=x^2-2x+3$

Differentiate the function with respect to x,

$$\begin{gathered} \begin{array}{rcl}& & f\text{'}\left(x\right)=2x-2 \\ \Rightarrow f\text{'}\left(c\right)=2c-2\end{array} \end{gathered}$$

Since,

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ \Rightarrow 2c-2=\frac{f\left({\displaystyle \frac{3}{2}}\right)-f\left(1\right)}{\left({\displaystyle \frac{3}{2}}-1\right)} \\ \Rightarrow 2c-2=\frac{{\displaystyle \frac{9}{4}}-3+3-1+2-1}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow 2c-2=\frac{{\left({\displaystyle \frac{3}{2}}\right)}^2-2\left({\displaystyle \frac{3}{2}}\right)+3-1+2-3}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow 2c-2=\frac{1}{2} \\ \Rightarrow 2c=\frac{5}{2} \\ \Rightarrow c=\frac{5}{4} \end{gathered}$$

Hence option $\left(B\right)$ is the correct option.