Question

Let $f\left(x\right)$ be a twice differentiable function for all real values of $x$ and satisfies $f\left(1\right)=1,f\left(2\right)=4,f\left(3\right)=9$. Then which of the following is definitely true?

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

Answer 1

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

Let $g\left(x\right)=f\left(x\right)-x^2$ is continuous and differentiable in given interval
Now, $g\left(1\right)=0,g\left(2\right)=0,g\left(3\right)=0[\because f(1)=1,f(2)=4,f(3)=9]$
From Rolle's theorem on $g\left(x\right):$
$g^{\prime }\left(x\right)=0$ for at least $x\in (1,2)$.
Let $g^{\prime }\left(c_1\right)=0$ where $c_1\in (1,2)$
similarly,
$g\left(x\right)=0$ for at least one $x\in (2,3)$.
Let $g^{\prime }\left(c_2\right)=0$ where $c_2\in (2,3)$
$\therefore g^{\prime }\left(c_1\right)=g^{\prime }\left(c_2\right)=0$
By rolle's theorem on $g^{\prime }\left(x\right):$
$g^{\prime \prime }\left(x\right)=0$ for at least one $x\in (c_1,c_2)$
$\Rightarrow f^{\prime \prime }\left(x\right)-2=0$
$\therefore f^{\prime \prime }\left(x\right)=2$ for some $x\in (1,3)$