Question

Let $f^{\prime }\left(x\right)=e^{x^2}$ and $f\left(0\right)=10.$ If $A<f\left(1\right)<B$ can be concluded from the mean value theorem, then the largest value of $(A-B)$ equals

• A. $e$
• B. $1-e$
• C. $e-1$
• D. $1+e$

Answer 1

The correct option is B $1-e$

Applying $\text{LMVT}$ in $[0,1]$ to the function $y=f\left(x\right)$, we get
$f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1},$ for some $c\in (0,1)$
$\Rightarrow e^{c^2}=\frac{f\left(1\right)-f\left(0\right)}{1}$
$\Rightarrow f\left(1\right)-10=e^{c^2}$ for some $c\in (0,1)$
But $1<e^{c^2}<e$ in $(0,1)$
$\Rightarrow 1<f\left(1\right)-10<e$
$\Rightarrow 11<f\left(1\right)<10+e$
$\Rightarrow A=11,B=10+e$
$\Rightarrow A-B=1-e$