JEE Advanced 2015 · Paper 2 · Mathematics Q46

← JEE Advanced 2015 · Paper 2

MathematicsInteger+4Question 46 of 60

If
$\alpha=\int_0^1\left(e^{9x+3\tan^{-1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$
where $\tan^{-1}x$ takes only principal values, then the value of $\left(\log_e|1+\alpha|-\dfrac{3\pi}4\right)$ is

Your answer

Answer and worked solution

Official answer: 9

Exact derivative.
$\frac d{dx}e^{9x+3\tan^{-1}x}=e^{9x+3\tan^{-1}x}\left(9+\frac3{1+x^2}\right)=e^{9x+3\tan^{-1}x}\cdot\frac{12+9x^2}{1+x^2}.$

Evaluate. $\alpha=e^{9+3\pi/4}-1$ , so
$\ln|1+\alpha|-\frac{3\pi}4=9.$

Answer: 9

Topics:Substitution Fundamental theorem of calculus

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JEE Advanced 2015 · Paper 2, Mathematics question 46