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

Answer 1

The correct option is C For every $x>0,$ there exists a $\beta \in (0,x)$ such that ${\psi }_2\left(x\right)=2x\left({\psi }_1\right(\beta )-1)$

$g\left(x\right)=\underset{0}{\overset{x^2}{\int }}\sqrt{t}{e}^{-t}dt,x>0$
Let $t=u^2\Rightarrow dt=2udu$
$\therefore g\left(x\right)=\underset{0}{\overset{x}{\int }}u{e}^{-u^2}\cdot 2udu=2\underset{0}{\overset{x}{\int }}{t}^2{e}^{-t^2}dt\dots \left(1\right)$
and
$f\left(x\right)=\underset{-x}{\overset{x}{\int }}(|t|-t^2)e^{-t^2}dt,x>0\therefore f\left(x\right)=2\underset{0}{\overset{x}{\int }}(t-t^2)e^{-t^2}dt\dots \left(2\right)$

From equation $\left(1\right)+\left(2\right),$ we get
$f\left(x\right)+g\left(x\right)=\underset{0}{\overset{x}{\int }}2t{e}^{-t^2}dt$
Let $t^2=p\Rightarrow 2tdt=dp$
$\therefore f\left(x\right)+g\left(x\right)=\underset{0}{\overset{x^2}{\int }}{e}^{-p}dp={[-e^{-p}]}_0^{x^2}\therefore f\left(x\right)+g\left(x\right)=1-e^{-x^2}\Rightarrow f\left(\sqrt{x}\right)+g\left(\sqrt{x}\right)=1-e^{-x}\Rightarrow f\left(\sqrt{\ln 3}\right)+g\left(\sqrt{\ln 3}\right)=1-e^{-\ln 3}=1-\frac{1}{3}=\frac{2}{3}$


From equation
$f\left(x\right)=\underset{-x}{\overset{x}{\int }}(|t|-t^2)e^{-t^2}dt,x>0\Rightarrow f\left(x\right)=2\underset{0}{\overset{x}{\int }}(|t|-t^2)e^{-t^2}dt\Rightarrow f^{\prime }\left(x\right)=2(x-x^2)e^{-x^2}\Rightarrow f^{\prime }\left(x\right)=-2x(x-1)e^{-x^2}$
$\therefore f\left(x\right)$ is increasing in $(0,1)$


${\psi }_1\left(x\right)=e^{-x}+x$
$\Rightarrow \psi {}_1^{\prime }\left(x\right)=1-e^{-x}<1,$ for $x>1$
Then for $\alpha \in (1,x),{\psi }_1\left(x\right)=1+\alpha x$ does not hold true for $\alpha >1$.


Now, ${\psi }_2\left(x\right)=x^2-2x-2e^{-x}+2$
$\Rightarrow {\psi }_2^{\prime }\left(x\right)=2x-2+2e^{-x}\Rightarrow {\psi }_2^{\prime }\left(x\right)=2{\psi }_1\left(x\right)-2$
From LMVT in $(0,x),$
$\frac{{\psi }_2\left(x\right)-{\psi }_2\left(0\right)}{x-0}=\psi {}_2^{\prime }\left(\beta \right)$
for $\beta \in (\infty ,x)$
$\Rightarrow {\psi }_2\left(x\right)=x\psi {}_2^{\prime }\left(\beta \right)\Rightarrow {\psi }_2\left(x\right)=2x\left({\psi }_1\right(\beta )-1)$