JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q9

MathematicsSingle correct+4 / -1Question 9 of 90

If $\displaystyle\int\frac{\left(x^2+1\right)e^x}{(x+1)^2}dx=f(x)e^x+C$ , where C is a constant, then $\dfrac{d^3f}{dx^3}$ at $x=1$ is equal to :

-\dfrac34

\dfrac34

-\dfrac32

\dfrac32

Answer and worked solution

Official answer: B

Form $\int(f+f')e^x=fe^x$ . Try $f=\dfrac{x-1}{x+1}$ : $f'=\dfrac2{(x+1)^2}$ and $f+f'=\dfrac{x^2+1}{(x+1)^2}$ ✓.

Third derivative. $f=1-2(x+1)^{-1}$ , $f'''=12(x+1)^{-4}$ , so at $x=1$ it is $\dfrac{12}{16}=\dfrac34$ .

(B)

Topics:Integration by parts Higher-order derivatives

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 9