Question

If $f\left(x\right)=x^2-2x+4$ and $\frac{f\left(5\right)-f\left(1\right)}{5-1}=f^{\prime }\left(c\right)$ then value of c will be

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is D 3

Given $f\left(x\right)=x^2-2x+4$

Replace $x$ by 5, we get,

$f\left(5\right)={\left(5\right)}^2-2\left(5\right)+4$

$\therefore f\left(5\right)=25-10+4$

$\therefore f\left(5\right)=19$

Now, replace $x$ by 1, we get,

$f\left(1\right)={\left(1\right)}^2-2\left(1\right)+4$

$\therefore f\left(1\right)=1-2+4$

$\therefore f\left(1\right)=3$

Given, $\frac{f\left(5\right)-f\left(1\right)}{5-1}=f^{\prime }\left(c\right)$

$\therefore \frac{19-3}{4}=f^{\prime }\left(c\right)$

$\therefore \frac{16}{4}=f^{\prime }\left(c\right)$ (1)

Now, $f^{\prime }\left(x\right)=2x-2$

$\therefore f^{\prime }\left(c\right)=2c-2$

Thus, from equation (1),

$4=2c-2$

$\therefore 2c=6$

$\therefore c=3$