Verified Solution • 2025–26 Session moderate

QJEEM23Ma42c0: JEE Main 2023 Mathematics - Class 12 JEEM23Ma42c0 Solution: For , if , where e = ∑ n = 0 ∞ 1 n ! and is constant of…

For α,β,γ,δ∈N\alpha, \beta, \gamma, \delta \in \mathbb{N} , if ∫((xe)2x+(ex)2x)log⁡exdx=1α(xe)βx−1γ(ex)δx+C\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=∑n=0∞1n! and C\mathrm{C} is constant of integration, then α+2β+3γ−4δ\alpha+2 \beta+3 \gamma-4 \delta is equal to :

Multiple Choice Options

  • A −8-8
  • B −4-4
  • C 1
  • D 4

Step-by-Step Verified Solution

We have,

∫((xe)2x+(ex)2x)log⁡exdx=1α(xe)βx−1γ(ex)δx+C\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

(xe)2xamp;=(elog⁡exe)2x[∵x=elog⁡ex]amp;=e2(xlog⁡ex−x).........(i)\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}

 Also, (ex)2x=(xe)−2x=e−2(xlog⁡ex−x)\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,

Iamp;=∫[(xe)2x+(ex)2x]log⁡exdxamp;=∫[e2(xlog⁡ex−x)+e−2(xlog⁡ex−x)]log⁡exdx     [From Eqs. (i) and (ii)] \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 xlog⁡ex−x=tx \log _e x-x=t

log⁡exdx=dt\log _e x d x=d t

Iamp;=∫(e2t+e−2t)dt=e2t2−e−2t2+Camp;=12(xe)2x−12(ex)2x+C\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 1α(xe)βx−1γ(ex)δx+C\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C

amp;α=2,β=2,γ=2,δ=2amp;∴α+2β+3γ−4δ=2+2×2+3×2−4×2=4\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring