Question

If $f\left(x\right)$ is a twice differentiable function for which $f\left(1\right)=1,f\left(2\right)=4$ and $f\left(3\right)=9$ then

• A. $f^{\prime \prime }\left(x\right)=2$ for all $xϵ(1,3)$
• B. $f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)=5$ for some $xϵ(2,3)$
• C. $f^{\prime \prime }\left(x\right)=3$ for all $xϵ(2,3)$
• D. $f^{\prime \prime }\left(x\right)=2$ for some $xϵ(1,3)$

Answer 1

The correct option is D $f^{\prime \prime }\left(x\right)=2$ for some $xϵ(1,3)$

Applying LMVT in the interval of $[1,2]$, give us

$f^{\prime }\left(a\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}=3$...(i) where $1<a<2$

In the interval of $[2,3]$

$f^{\prime }\left(b\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}=5$ where $2<b<3$

Since f(x) is a twice differentiable function, thus we can apply LMVT on the first derivative of f(x).

Hence in the interval of $[a,b]$

$f"\left(c\right)=\frac{f^{\prime }\left(b\right)-f^{\prime }\left(a\right)}{b-a}$

$=\frac{5-3}{b-a}$

$=\frac{2}{b-a}$

Considering the difference of b and a be 1, we get

$f"\left(c\right)=2$

Since $1<a<2$ and $2<b<3$ hence $1<c<3$

Or

$cϵ(1,3)$.