Mathematics - Application of Derivative Question with Solution | TestHub

(A) Given

 $f\left(x\right)=2{\int }_0^x\left(t-t^2\right)e^{-t^2}dt;x>0$

$g\left(x\right)={\int }_0^{x^2}\sqrt{t}{\mathrm{e}}^{-1}dt;x>0$
put $t=u^2$
$\therefore g\left(x\right)=2{\int }_0^x{u}^2{e}^{-u^2}du=2{\int }_0^x{t}^2{e}^{-t^2}dt$

now, $f\left(x\right)+g\left(x\right)={\int }_0^{x}2t{\mathrm{e}}^{-t^2}dt=1-{\mathrm{e}}^{-x^2}$

$\Rightarrow f\left(\sqrt{\ln 3}\right)+\mathrm{g}\left(\sqrt{\ln 3}\right)=1-e^{-\ln 3}=\frac{2}{3}$

(B) for $\alpha \in (1,x),$ ${\psi }_1\left(x\right)=1+\alpha x$

$\Rightarrow e^{-x}+x-1-\alpha x=0$

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

(C) ${\psi }_2\left(x\right)=2x\left({\psi }_1\left(\beta \right)-1\right)$

${{\psi }^{\text{'}}}_2\left(x\right)=2{\psi }_1\left(x\right)-2$

from LMVT, $\frac{{\psi }_2\left(\mathrm{x}\right)-{\psi }_2\left(0\right)}{\mathrm{x}-0}={\psi }_2^{\text{'}}\left(\beta \right)$ for atleast one $\beta \in (0,x)$

$\Rightarrow {\psi }_2\left(\mathrm{x}\right)=2\mathrm{x}\left({\psi }_1\left(\beta \right)-1\right)$

(D) Given

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

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

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

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

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