Paragraph: Let _ 1 :[0, ∞) arrow ℝ, _ 2 :[0, ∞) arrow ℝ, f:[0, ∞)...

(A) Given

$f\left(x\right) = 2{∫}_{0}^{x}\left(t - {t}^{2}\right){e}^{- {t}^{2}}dt; x > 0$

$g\left(x\right) = {∫}_{0}^{{x}^{2}}\sqrt{t}{e}^{- 1}dt; x > 0$
put $t = {u}^{2}$
$∴ g\left(x\right) = 2{∫}_{0}^{x}{u}^{2}{e}^{- {u}^{2}}du = 2{∫}_{0}^{x}{t}^{2}{e}^{- {t}^{2}}dt$

now, $f\left(x\right) + g\left(x\right) = {∫}_{0}^{x}2t{e}^{- {t}^{2}}dt = 1 - {e}^{- {x}^{2}}$

$⇒ f ( \sqrt{\ln 3} ) + g ( \sqrt{\ln 3} ) = 1 - {e}^{- \ln 3} = \dfrac{2}{3}$

(B) for $\alpha \in ( 1, x ) ,$ ${ψ}_{1} ( x ) = 1 + \alphax$

$⇒ {e}^{- x} + x - 1 - \alphax = 0$

$⇒ \left({e}^{- x} - 1\right) = x ( \alpha - 1 )$
Which is not possible because LHS $< 0$ & RHS $> 0$

(C) ${ψ}_{2}$\left(x\right) = 2x$\left({ψ}_{1}\left(\beta\right) - 1\right)$

${ψ}^{'}2 ( x ) = 2{ψ}_{1} ( x ) - 2$

from LMVT, $\dfrac{{ψ}_{2} ( x ) - {ψ}_{2} ( 0 )}{x - 0} = {ψ}_{2}^{'}\left(\beta\right)$ for atleast one $\beta \in ( 0, x )$

$⇒ {ψ}_{2} ( x ) = 2x\left({ψ}_{1} ( \beta ) - 1\right)$

(D) Given

$f\left(x\right) = {∫}_{- x}^{x}\left(\left|t\right| - {t}^{2}\right){e}^{- {t}^{2}}dt, x > 0$

${f}^{'}\left(x\right) = 2\left(x - {x}^{2}\right){e}^{- {x}^{2}}$

${f}^{'}\left(x\right) = 2x\left(1 - x\right){e}^{- {x}^{2}}$

${f}^{'}\left(x\right) \ge 0\text{for}x \in \left[0, 1\right]$ So, increasing in this interval.

$\text{and}{f}^{'}\left(x\right) \le 0\text{for}x \in [ 1, \infty )$ So, decreasing in this interval.