JEE Main 2023 — Mathematics Calculus

Given:

$$f(x)+{\int }_{3}^{x}\frac{f(t)}{t}dt=\sqrt{x+1}$$f(x)+∫3x​tf(t)​dt=x+1​

$$\Rightarrow {f}^{'}(x)+\frac{f(x)}{x}=\frac{1}{2\sqrt{x+1}}$$⇒f′(x)+xf(x)​=2x+1​1​

Put $y=f(x)$y=f(x), then

$$\frac{dy}{dx}+\frac{y}{x}=\frac{1}{2\sqrt{x+1}}$$dxdy​+xy​=2x+1​1​

So, $\displaystyle I.F.={e}^{\int \frac{dx}{x}}={e}^{\mathrm{ln}|x|}=x$I.F.=e∫xdx​=eln∣x∣=x

Hence, solution is

$$xy=\frac{1}{2}\int \frac{x}{\sqrt{x+1}}dx$$xy=21​∫x+1​x​dx

$$\Rightarrow xy=\frac{1}{2}\int (\frac{x+1-1}{\sqrt{x+1}})dx$$⇒xy=21​∫(x+1​x+1−1​)dx

$$\Rightarrow xy=\frac{1}{2}\int (\sqrt{x+1}-\frac{1}{\sqrt{x+1}})dx$$⇒xy=21​∫(x+1​−x+1​1​)dx

$$\Rightarrow xy=\frac{1}{2}(\frac{2}{3}{(x+1)}^{\frac{3}{2}}-2\sqrt{x+1})+C$$⇒xy=21​(32​(x+1)23​−2x+1​)+C

$$\Rightarrow xy=\frac{1}{3}{(x+1)}^{\frac{3}{2}}-\sqrt{x+1}+C$$⇒xy=31​(x+1)23​−x+1​+C

Put $x=3$x=3, then $f(3)=2$f(3)=2

So,

$$6=\frac{8}{3}-2+C\Rightarrow C=\frac{16}{3}$$6=38​−2+C⇒C=316​

Hence,

$$xy=\frac{1}{3}{(x+1)}^{\frac{3}{2}}-\sqrt{x+1}+\frac{16}{3}$$xy=31​(x+1)23​−x+1​+316​

So, put $x=8$x=8, then

$$8f(8)=\frac{27}{3}-3+\frac{16}{3}$$8f(8)=327​−3+316​

$$\Rightarrow f(8)=\frac{34}{3\times 8}$$⇒f(8)=3×834​

$$\Rightarrow 12f(8)=17$$⇒12f(8)=17