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 $f\left(0\right)=g\left(0\right)=0,$
${\psi }_1\left(x\right)=e^{-x}+x,x\ge 0$
${\psi }_2\left(x\right)=x^2-2x-2e^{-x}+2,x\ge 0$
$f\left(x\right)=\underset{-x}{\overset{x}{\int }}(|t|-t^2)e^{-t^2}dt,x>0$
and $g\left(x\right)=\underset{0}{\overset{x^2}{\int }}\sqrt{t}{e}^{-t}dt,x>0$

Which of the following statements is TRUE?

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

Answer 1

The correct option is D $g\left(x\right)\le \frac{2}{3}{x}^3-\frac{2}{5}{x}^5+\frac{1}{7}{x}^7,$ for all $x\in (0,\frac{1}{2})$

${\psi }_1\left(x\right)=e^{-x}+x$
$\Rightarrow \psi {}_1^{\prime }\left(x\right)=1-e^{-x}>0,$ for $x>0$
and ${\psi }_1\left(x\right)\in (1,\infty )$

${\psi }_2\left(x\right)=x^2-2x-2e^{-x}+2$ for $x>0$
$\Rightarrow \psi {}_2^{\prime }\left(x\right)=2x-2+2e^{-x}=2(x+e^{-x})-2$
$\because x+e^{-x}\in (1,\infty )$ from above part
$\therefore 2(x+e^{-x})\in (2,\infty )$
$\Rightarrow 2(x+e^{-x})-2\in (0,\infty )$
$\Rightarrow \psi {}_2^{\prime }\left(x\right)>0$
and ${\psi }_2\left(x\right)\in \left({\psi }_2\right(0),{\psi }_2(x\to \infty \left)\right)$
$\Rightarrow {\psi }_2\left(x\right)\in (0,\infty )$ for $x>0$


and
$f\left(x\right)={\int }_{-x}^x(|t|-t^2)e^{-t^2}dt=2{\int }_0^x(t-t^2)e^{-t^2}dt={\int }_0^{x}2t{e}^{-t^2}dt-2{\int }_0^x{t}^2{e}^{-t^2}dt={\int }_0^{x}2t{e}^{-t^2}dt-2{\int }_0^x{t}^2(1-\frac{t^2}{1!}+\frac{t^4}{2!}-\frac{t^6}{3!}+\dots )dt\therefore f\left(x\right)\le 1-e^{-x^2}-2{\int }_0^x{t}^2(1-\frac{t^2}{1!})dt\Rightarrow f\left(x\right)\le 1-e^{-x^2}-\frac{2}{3}{x}^3+\frac{2}{5}{x}^5$ for all $x\in (0,\frac{1}{2})$


Now, $\sqrt{t}{e}^{-t}=\sqrt{t}(1-\frac{t}{1!}+\frac{t^2}{2!}-\frac{t^3}{3!}+\dots ,\infty )$
and $\sqrt{t}{e}^{-t}\le t^{1/2}-t^{3/2}+\frac{1}{2}{t}^{5/2}$
$\therefore {\int }_0^{x^2}\sqrt{t}{e}^{-t}dt\le {\int }_0^{x^2}(t^{1/2}-t^{3/2}+\frac{1}{2}{t}^{5/2})dt\Rightarrow g\left(x\right)\le \frac{2}{3}{x}^3-\frac{2}{5}{x}^5+\frac{1}{7}{x}^7$ for all $x\in (0,\frac{1}{2})$