Question

Let ${\Psi }_1:[0,\infty )\to R,{\Psi }_2:[0,\infty )\to R,f:[0,\infty )\to R$ and $g:[0,\infty )\to R$ be functions such that $f(0)=g(0)=0$,
${\psi }_1(x)=e^{-x}+x,x\ge 0,$
${\psi }_2(x)=x^2-2x-2e^{-x}+2,x\ge 0,$
$f(x)=\underset{-x}{\overset{x}{\int }}\left(|t|-t^2\right)e^{-t^2}dt,x>0$
and $g(x)=\underset{0}{\overset{x^2}{\int }}\sqrt{t}{e}^{-t}dt,x>0$.
Which of the following statements is TRUE?

• A. $f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}$
• B. For every $x>1$, there exists and $\alpha \in (1,x)$ such that ${\psi }_1(x)=1+\alpha x$
• C. For every $x>0$, there exists a $\beta \in (0,x)$ such that ${\psi }_2(x)=2x\left({\psi }_1(\beta )-1\right)$
• D. $f$ is an increasing function on the interval $\left[0,\frac{3}{2}\right]$

Answer 1

Answer: (C)

${\psi }_2(x)=x^2-2x-2e^{-x}+2,x\ge 0$
${\psi }_2^{\prime }(x)=2x-2+2e^{-x}=2\left(x+e^{-x}-1\right)$
${\psi }_2^{\prime }(\beta )=2\left({\psi }_1(\beta )-1\right)$
Since ${\psi }_2(x)$ is a continuous and differentiable function $\forall x\in [0,x]$
${\psi }_2(0)=0,{\psi }_2(x)=x^2-2x-2e^{-x}+2$
Hence according to LMVT there exist at least one $\beta \in (0,x)$ such that
$\left(\frac{{\psi }_2(x)-{\psi }_2(0)}{x}\right)={\psi }_2^{\prime }(\beta )$
$=\frac{{\psi }_2(x)}{x}=2\left({\psi }_1(\beta )-1\right)$
$={\psi }_2(x)=2x\left({\psi }_1(\beta )-1\right)$