JEE Main 2022 — Mathematics Calculus

$$\int \frac{({x}^{2}+1){e}^{x}dx}{{(x+1)}^{2}}=f(x){e}^{x}+C$$∫(x+1)2(x2+1)exdx​=f(x)ex+C

$$\int {e}^{x}(\frac{{x}^{2}-1}{{(x+1)}^{2}}+\frac{2}{{(x+1)}^{2}})dx=f(x){e}^{x}+C$$∫ex((x+1)2x2−1​+(x+1)22​)dx=f(x)ex+C

$$\int {e}^{x}(\frac{x-1}{x+1}+\frac{2}{{(x+1)}^{2}})dx=f(x){e}^{x}+C$$∫ex(x+1x−1​+(x+1)22​)dx=f(x)ex+C

We know that $\displaystyle \int {e}^{x}(f(x)+{f}^{'}(x))={e}^{x}f(x)+c$∫ex(f(x)+f′(x))=exf(x)+c

Here f(x)=\frac{x-1}{x+1}&{f}^{'}(x)=\frac{2}{{(x+1)}^{2}}

So $\displaystyle \int {e}^{x}((\frac{x-1}{x+1})+\frac{2}{(x+1{)}^{2}})dx=f(x){e}^{x}+C$∫ex((x+1x−1​)+(x+1)22​)dx=f(x)ex+C

$$\Rightarrow {e}^{x}(\frac{x-1}{x+1})+C={e}^{x}f(x)+C$$⇒ex(x+1x−1​)+C=exf(x)+C

On comparing both sides we get $f(x)=\frac{x-1}{x+1}$f(x)=x+1x−1​

So {f}^{'}(x)=\frac{2}{{(x+1)}^{2}}&{f}^{"}(x)=\frac{-4}{{(x+1)}^{3}}

${f}^{'''}(x)=\frac{12}{{(x+1)}^{4}}$f′′′(x)=(x+1)412​ ,

Now ${f}^{'''}(1)=\frac{12}{{(1+1)}^{4}}=\frac{12}{16}=\frac{3}{4}$f′′′(1)=(1+1)412​=1612​=43​