JEE Mathematics 2023 Question with Solution

Given:

$$\int \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \log x \, dx = \frac{1}{\alpha} \left( \frac{x}{e} \right)^{\beta x} - \frac{1}{\gamma} \left( \frac{e}{x} \right)^{\delta x} + C$$

Find: $\alpha + 2\beta + 3\gamma - 4\delta$

Differentiate the right-hand side with respect to $x$.

Let

$$y = \left(\frac{x}{e}\right)^{\beta x}$$

Then

$$\ln y = \beta x (\ln x - 1)$$

So,

$$\frac{1}{y} \frac{dy}{dx} = \beta (\ln x - 1) + \beta x \cdot \frac{1}{x} = \beta \ln x$$

Hence,

$$\frac{dy}{dx} = \beta \left(\frac{x}{e}\right)^{\beta x} \ln x$$

Now let

$$z = \left(\frac{e}{x}\right)^{\delta x}$$

Then

$$\ln z = \delta x (1 - \ln x)$$

So,

$$\frac{1}{z} \frac{dz}{dx} = \delta (1 - \ln x) - \delta = -\delta \ln x$$

Hence,

$$\frac{dz}{dx} = -\delta \left(\frac{e}{x}\right)^{\delta x} \ln x$$

Therefore, the derivative of the right-hand side 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 as extracted in the solution, we get

$$\beta x = 2x \implies \beta = 2$$ $$\alpha = \beta \implies \alpha = 2$$ $$\delta x = 2x \implies \delta = 2$$ $$\gamma = \delta \implies \gamma = 2$$

Now,

$$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 4$$

Therefore, the numerical result is $4$. The solution explicitly marks the correct option as D, which disagrees with the listed options.

Given: an antiderivative in terms of $\alpha, \beta, \gamma, \delta$.

Find: the value of $\alpha + 2\beta + 3\gamma - 4\delta$.

The method is to differentiate the proposed antiderivative and match its derivative with the integrand.

For

$$\left(\frac{x}{e}\right)^{\beta x}$$

write logarithmically:

$$\ln y = \beta x (\ln x - 1)$$

Differentiating,

$$\frac{1}{y}\frac{dy}{dx} = \beta \ln x$$

so

$$\frac{dy}{dx} = \beta \left(\frac{x}{e}\right)^{\beta x} \ln x$$

Thus,

$$\frac{d}{dx}\left[\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}\right] = \frac{\beta}{\alpha}\left(\frac{x}{e}\right)^{\beta x} \ln x$$

Similarly, for

$$\left(\frac{e}{x}\right)^{\delta x}$$

let

$$\ln z = \delta x(1 - \ln x)$$

Then

$$\frac{1}{z}\frac{dz}{dx} = -\delta \ln x$$

so

$$\frac{dz}{dx} = -\delta \left(\frac{e}{x}\right)^{\delta x} \ln x$$

Hence,

$$\frac{d}{dx}\left[-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}\right] = \frac{\delta}{\gamma}\left(\frac{e}{x}\right)^{\delta x} \ln x$$

Matching powers and coefficients from the solution gives

$$\beta = 2, \quad \alpha = 2, \quad \delta = 2, \quad \gamma = 2$$

Therefore,

$$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 4 + 6 - 8 = 4$$

So the computed value is $4$.