Question

For, α, β, γ, δ ∈ \(\mathbb{N},\) if \(∫\left((\frac{x}{e})^{2x}+(\frac{e}{x})^{2x}\right)log_e\,xdx=\) \(\frac{1}{\alpha}(\frac{x}{e})^{βx}-\frac{1}{γ}(\frac{e}{x})^{δx}+c,\) Where \(e=\displaystyle\sum_{n=10}^{∞}\frac{1}{n!}\) and C is constant of integration, then α + 2β + 3γ - 4δ is equal to

• A. -8
• B. -4
• C. 1
• D. 4

Answer 1

The Correct Option is D

Let's differentiate the right-hand side with respect to $x$: $\frac{d}{dx} \left[ \frac{1}{\alpha} \left(\frac{x}{e}\right)^{\beta x} - \frac{1}{\gamma} \left(\frac{e}{x}\right)^{\delta x} + C \right]$ 
Let $y = \left(\frac{x}{e}\right)^{\beta x}$. Then $\ln y = \beta x (\ln x - 1)$. $\frac{1}{y} \frac{dy}{dx} = \beta (\ln x - 1) + \beta x \cdot \frac{1}{x} = \beta \ln x$. $\frac{dy}{dx} = \beta \left(\frac{x}{e}\right)^{\beta x} \ln x$. 
Let $z = \left(\frac{e}{x}\right)^{\delta x}$. Then $\ln z = \delta x (1 - \ln x)$. $\frac{1}{z} \frac{dz}{dx} = \delta (1 - \ln x) - \delta = -\delta \ln x$. $\frac{dz}{dx} = -\delta \left(\frac{e}{x}\right)^{\delta x} \ln x$. 
Therefore, the derivative is: $\frac{\beta}{\alpha} \left(\frac{x}{e}\right)^{\beta x} \ln x + \frac{\delta}{\gamma} \left(\frac{e}{x}\right)^{\delta x} \ln x$
Comparing with the integrand, we have: 
$\beta x = 2x \implies \beta = 2$ $\alpha = \beta \implies \alpha = 2$ $\delta x = 2x \implies \delta = 2$ $\gamma = \delta \implies \gamma = 2$ $\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 2 + 4 + 6 - 8 = 4$ 
Answer: 4.