Question

If the mean value theorem is $f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(c\right)$. 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 for the correct option:

Mean value theorem:

$f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(c\right)$

$\Rightarrow$$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ …$\left[1\right]$

The given function is

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

Differentiating the function with respect to $x$ we get

$\Rightarrow$$f\text{'}\left(x\right)=2x-2$

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

For the given interval $\left[1,\frac{3}{2}\right]$ we have $a=1,b=\frac{3}{2}$

$f\left(a\right)=f\left(1\right)={\left(1\right)}^2-2\left(1\right)+3$

$\Rightarrow$$f\left(a\right)=2$

$f\left(b\right)=f\left(\frac{3}{2}\right)={\left(\frac{3}{2}\right)}^2-2\left(\frac{3}{2}\right)+3$

$\Rightarrow$$f\left(b\right)=\frac{9}{4}$

Substituting all these required values in equation $\left[1\right]$

$\Rightarrow$ $2c-2=\frac{{\displaystyle \frac{9}{4}}-2}{{\displaystyle \frac{3}{2}}-1}$

$\Rightarrow$$2\left(c-1\right)=\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{2}}}$

$\Rightarrow$ $\left(c-1\right)=\frac{1}{4}$

$\Rightarrow$ $c=\frac{5}{4}$

Thus the required value of $c$ is $\frac{5}{4}$.

Hence the correct option is option(B)