JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 7)
For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
, where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :
$$-8$$
$$-4$$
1
4
Explanation
We have,
$$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
$$ \begin{aligned} \left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\right)^{2 x} \left[\because x=e^{\log _e x}\right] \\\\ & =e^{2\left(x \log _e x-x\right)} .........(i) \end{aligned} $$
$$ \text { Also, }\left(\frac{e}{x}\right)^{2 x}=\left(\frac{x}{e}\right)^{-2 x}=e^{-2\left(x \log _e x-x\right)} $$ .........(ii)
Let,
$$ \begin{aligned} I & =\int\left[\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right] \log _e x d x \\\\ & =\int\left[e^{2\left(x \log _e x-x\right)}+e^{-2\left(x \log _e x-x\right)}\right] \log _e x d x ~~~~\text{ [From Eqs. (i) and (ii)] } \end{aligned} $$
Let $x \log _e x-x=t$
$$ \log _e x d x=d t $$
$$ \begin{aligned} I & =\int\left(e^{2 t}+e^{-2 t}\right) d t=\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C \\\\ & =\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C \end{aligned} $$
On comparing I with $\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$
$$ \begin{aligned} & \alpha=2, \beta=2, \gamma=2, \delta=2 \\\\ & \therefore \alpha+2 \beta+3 \gamma-4 \delta=2+2 \times 2+3 \times 2-4 \times 2=4 \end{aligned} $$
$$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
$$ \begin{aligned} \left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\right)^{2 x} \left[\because x=e^{\log _e x}\right] \\\\ & =e^{2\left(x \log _e x-x\right)} .........(i) \end{aligned} $$
$$ \text { Also, }\left(\frac{e}{x}\right)^{2 x}=\left(\frac{x}{e}\right)^{-2 x}=e^{-2\left(x \log _e x-x\right)} $$ .........(ii)
Let,
$$ \begin{aligned} I & =\int\left[\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right] \log _e x d x \\\\ & =\int\left[e^{2\left(x \log _e x-x\right)}+e^{-2\left(x \log _e x-x\right)}\right] \log _e x d x ~~~~\text{ [From Eqs. (i) and (ii)] } \end{aligned} $$
Let $x \log _e x-x=t$
$$ \log _e x d x=d t $$
$$ \begin{aligned} I & =\int\left(e^{2 t}+e^{-2 t}\right) d t=\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C \\\\ & =\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C \end{aligned} $$
On comparing I with $\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$
$$ \begin{aligned} & \alpha=2, \beta=2, \gamma=2, \delta=2 \\\\ & \therefore \alpha+2 \beta+3 \gamma-4 \delta=2+2 \times 2+3 \times 2-4 \times 2=4 \end{aligned} $$
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