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. ${\psi }_1(x)\le 1$, for all $x>0$
• B. ${\psi }_2(x)\le 0$, for all $x>0$
• C. $f(x)\ge 1-e^{-x^2}-\frac{2}{3}{x}^3+\frac{2}{5}{x}^5$, for all $x\in \left(0,\frac{1}{2}\right)$
• D. $g(x)\le \frac{2}{3}{x}^3-\frac{2}{5}{x}^5+\frac{1}{7}{x}^7$, for all $x\in \left(0,\frac{1}{2}\right)$

Answer 1

Answer: (D)

${\psi }^{\prime }{}_1(x)=1-e^{-x}>0$
${\psi }_2(x)=(x-1{)}^2+1-2e^{-x}>0$ for $x=1$
$\because e^{-t}=1-t+\frac{t^2}{2}-\frac{t^3}{3}+\dots$
$\sqrt{t}{e}^{-t}=\sqrt{t}-t^{3/2}+\frac{1}{2}{t}^{5/2}\dots$
So, $\sqrt{t}{e}^{-t}<\sqrt{t}-t^{3/2}+\frac{1}{2}{t}^{5/2}$ for $t\in (0,1)$
$\underset{0}{\overset{x^2}{\int }}\sqrt{t}{e}^{-t}dt<\underset{0}{\overset{x^2}{\int }}\left(\sqrt{t}-t^{3/2}+\frac{1}{2}{t}^{5/2}\right)dt$
$=\frac{2}{3}{x}^3-\frac{2}{5}{x}^5+\frac{1}{7}{x}^7$