JEE Main 2024 — Mathematics Calculus

Given: $\displaystyle g(x)={\int }_{0}^{{x}^{2}}{t}^{\frac{1}{2}}{e}^{-t}dt$g(x)=∫0x2​t21​e−tdt

Let, $t={y}^{2}$t=y2

$$\Rightarrow dt=2ydy$$⇒dt=2ydy

$$\Rightarrow g(x)={\int }_{0}^{x}2{y}^{2}{e}^{-{y}^{2}}dy$$⇒g(x)=∫0x​2y2e−y2dy

$$\Rightarrow g(x)=2{\int }_{0}^{x}{t}^{2}{e}^{-{t}^{2}}dt$$⇒g(x)=2∫0x​t2e−t2dt

Also, $\displaystyle f(x)=2{\int }_{0}^{x}(t-{t}^{2}){e}^{-{t}^{2}}dt$f(x)=2∫0x​(t−t2)e−t2dty

$$\Rightarrow f(x)+g(x)=2{\int }_{0}^{x}t{e}^{-{t}^{2}}dt$$⇒f(x)+g(x)=2∫0x​te−t2dt

Let, ${t}^{2}=p$t2=p

$$\Rightarrow 2tdt=dp$$⇒2tdt=dp

$$\Rightarrow f(x)+g(x)={\int }_{0}^{{x}^{2}}{e}^{-p}dp$$⇒f(x)+g(x)=∫0x2​e−pdp

$$\Rightarrow f(x)+g(x)=1-{e}^{-{x}^{2}}$$⇒f(x)+g(x)=1−e−x2

$$\Rightarrow 9f(\sqrt{{\mathrm{log}}_{e}9})+g(\sqrt{{\mathrm{log}}_{e}9})=9(1-{e}^{-\mathrm{log}9})$$⇒9f(loge​9​)+g(loge​9​)=9(1−e−log9)

$$\Rightarrow 9f(\sqrt{{\mathrm{log}}_{e}9})+g(\sqrt{{\mathrm{log}}_{e}9})=9(1-\frac{1}{9})$$⇒9f(loge​9​)+g(loge​9​)=9(1−91​)

$$\Rightarrow 9f(\sqrt{{\mathrm{log}}_{e}9})+g(\sqrt{{\mathrm{log}}_{e}9})=8$$⇒9f(loge​9​)+g(loge​9​)=8