Question

For every twice differentiable function $f:R\to [-2,2]$ with $\left(f\right(0){)}^2+(f^{\prime }\left(0\right){)}^2=85$, which of the following statement(s) is (are) TRUE?

• A. There exist $r,s\in R$, where $r<s$, such that $f$ is one-one on the open interval $(r,s)$
• B. There exists $x_0\in (-4,0)$ such that $|f^{\prime }\left(x_0\right)|\le 1$
• C. $\underset{x\to \infty }{lim}f\left(x\right)=1$
• D. There exists $\alpha \in (-4,4)$ such that $f\left(\alpha \right)+f^{\prime \prime }\left(\alpha \right)=0$ and $f^{\prime }\left(\alpha \right)\ne 0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A There exist $r,s\in R$, where $r<s$, such that $f$ is one-one on the open interval $(r,s)$
B There exists $x_0\in (-4,0)$ such that $|f^{\prime }\left(x_0\right)|\le 1$
D There exists $\alpha \in (-4,4)$ such that $f\left(\alpha \right)+f^{\prime \prime }\left(\alpha \right)=0$ and $f^{\prime }\left(\alpha \right)\ne 0$

$\left(f\right(0){)}^2+(f^{\prime }\left(0\right){)}^2=85$
$f$ can't be constant as if $f$ is constant, then $f^{\prime }\left(c\right)=0\text{and}f\left(c\right)=\sqrt{85}\notin R_f,\forall c\in R$
$\therefore$ Option 1 is true for every continuos function as we can find an interval $(r,s)$ with $r<s$ such that $f$ is monotone.

Consider $f:[-4,0]\to [-2,2]$
Applying MVT, we get
$f^{\prime }\left(x_0\right)=\frac{f\left(0\right)-f(-4)}{4}$
$\Rightarrow |f^{\prime }\left(x_0\right)|\le \frac{1}{4}|f(-4)|+|f\left(0\right)|$
$\le \frac{2+2}{4}=1,x_0\in (-4,0)$

If $\underset{x\to \infty }{lim}f\left(x\right)=1$,
then $y=1$ is the horizontal asymptote.
$\Rightarrow \underset{x\to \infty }{lim}{f}^{\prime }\left(x\right)=0$
which is not possible as proved earlier.
$\therefore$ Option 3 is wrong.

Consider $g\left(x\right)=\left(f\right(x){)}^2+(f^{\prime }\left(x\right){)}^2$
$\text{then}g\left(0\right)=85$
Applying MVT on $f$ in the interval $[-4,0]$,
$\exists x_1\in (-4,0)\text{such that}{f}^{\prime }\left(x_1\right)\le 1$
Also, $f\left(x\right)\le 2$
$\Rightarrow \exists x_1\in (-4,0)\text{such that}g\left(x_1\right)\le 5$

$\text{Similarly},\exists x_2\in (0,4)\text{such that}g\left(x_2\right)\le 5$
Since, $g$ is continuous, we can conclude from the above results that $g^{\prime }\left(x\right)=0$ for some $x\in (-4,4)$.

$g^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$
$=2f^{\prime }\left(x\right)\left[f\right(x)+f^{\prime \prime }(x\left)\right]$
$\therefore$ There exists $\alpha \in (-4,4)$ such that $f\left(\alpha \right)+f^{\prime \prime }\left(\alpha \right)=0$ and $f^{\prime }\left(\alpha \right)\ne 0$