Mathematics - Application of Derivative Question with Solution | TestHub

(A) Given

${\psi }_1\left(x\right)=e^{-x}+x, x\ge 0$

${{\psi }^{\text{'}}}_1\left(x\right)=-e^{-x}+1, x\ge 0$

${{\psi }^{\text{'}}}_1\left(x\right)$ is increasing function.

${\psi }_1\left(x\right)>{\psi }_1\left(0\right) \forall x\ge 0$

${\psi }_1\left(x\right)\ge 1$

So,

$e^{-x}+x<1$ for $x\in (0,\infty )$ is incorrect.

LHS is increasing and unbounded function.

(B) Given${\psi }_2\left(x\right)=x^2-2x-2e^{-x}+2, x\ge 0$

${{\psi }^{\text{'}}}_2\left(x\right)=2x-2+2e^{-x}, x\ge 0$

${{\psi }^{\text{'}}}_2\left(x\right)=2{\psi }_1\left(x\right)-2 \ge 0, x\ge 0$

So, ${\psi }_2\left(x\right)$ is increasing function.

$x^2-2x-2e^{-x}+2<1$
for $x\in (0,\infty )$ is incorrect because LHS $\to \infty$ when $x\to \infty$

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

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

$=1-e^{-x^2}-2{\int }_0^x{t}^2\left(1-t^2+\frac{t^4}{2!}\cdots ..\right)$

$=f\left(x\right)-1+e^{-x^2}+\frac{2x^3}{3}-\frac{2x^5}{5}<0$ in $\left(0,\frac{1}{2}\right)$

$g\left(x\right)={\int }_0^{x^2}\sqrt{t}{e}^{-1}dt$

Put $t=z^2$

$g\left(x\right)={\int }_0^{x}2z^2{e}^{-z^2}dz$

$={\int }_0^{x}2z^2\left(1-z^2+\frac{z^4}{2!}\dots ..\right)dz$

$g\left(x\right)=\frac{2x^3}{3}-\frac{2x^5}{5}+\frac{2x^7}{2.7}-\frac{2·x^9}{9.6}-\dots \dots$

$g\left(x\right)-\frac{2x^3}{3}+\frac{2x^5}{5}-\frac{2x^7}{2.7}\le 0$

$=1-e^{-x^2}-\frac{2x^3}{3}+\frac{2x^5}{5}-\frac{2x^7}{2}+\frac{2x^9}{6}\dots \dots \dots$

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

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

$\Rightarrow f\left(x\right)\le 1-e^{-x^2}-\left(\frac{2}{3}{x}^3-\frac{2}{3}{x}^5\right)$

(D) $g\left(x\right)={\int }_0^{x^2}\sqrt{t}{e}^{-t}dt$

$\Rightarrow g\left(x\right)\le {\int }_0^{x^2}\sqrt{t}\left(1-t+\frac{t^2}{2}\right)dt\Rightarrow g\left(x\right)\le \frac{2}{3}{x}^3-\frac{2}{5}{x}^5+\frac{1}{7}{x}^7$

$g\left(x\right)\ge {\int }_0^{x^2}\sqrt{t}\left(1-t\right)dt\Rightarrow g\left(x\right)\ge \frac{2}{3}{x}^3-\frac{2}{5}{x}^5$