JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q25

← JEE Main 2022 · 30 Jun · Shift 1

MathematicsInteger+4 / -1Question 25 of 90

Let $f(t)=\displaystyle\int_0^te^{x^3}\left(\frac{x^8}{\left(x^6+2x^3+2\right)^2}\right)dx$ . If $f(1)+f'(1)=\alpha e-\dfrac16$ , then the value of $150\alpha$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 16

Antiderivative. $\dfrac d{dx}\left[\dfrac{e^{x^3}}{3(x^6+2x^3+2)}\right]=\dfrac{x^8e^{x^3}}{(x^6+2x^3+2)^2}$ , so $f(1)=\dfrac e{15}-\dfrac16$ .

Derivative. $f'(1)=\dfrac e{25}$ .

Sum. $\alpha=\frac1{15}+\frac1{25}=\frac8{75}$ , so $150\alpha=16$ .

Answer: 16

Topics:Fundamental theorem of calculus Substitution

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 25