Integral Equation Involving Logarithmic Substitution and Parameter Solution

To determine which of the statements are true, we start by evaluating the given definite integral:

$I = \int_{1}^{e} \frac{(\log_e x)^{1/2}}{x\left(a - (\log_e x)^{3/2}\right)^2} dx$

Let us use the substitution: $u = (\log_e x)^{3/2}$

Differentiating both sides with respect to $x$, we get: $du = \frac{3}{2} (\log_e x)^{1/2} \cdot \frac{1}{x} \, dx \implies \frac{(\log_e x)^{1/2}}{x} \, dx = \frac{2}{3} \, du$

Next, we change the limits of integration according to the substitution:

• When $x = 1$, $u = (\log_e 1)^{3/2} = 0$.
• When $x = e$, $u = (\log_e e)^{3/2} = 1^{3/2} = 1$.

Substituting these into the integral gives: $I = \int_{0}^{1} \frac{\frac{2}{3} \, du}{(a - u)^2} = \frac{2}{3} \int_{0}^{1} (a - u)^{-2} \, du$

Evaluating the integral: $I = \frac{2}{3} \left[ \frac{1}{a - u} \right]_{0}^{1} = \frac{2}{3} \left( \frac{1}{a - 1} - \frac{1}{a} \right) = \frac{2}{3} \left( \frac{a - (a - 1)}{a(a - 1)} \right) = \frac{2}{3a(a - 1)}$

We are given that $I = 1$: $\frac{2}{3a(a - 1)} = 1 \implies 3a(a - 1) = 2 \implies 3a^2 - 3a - 2 = 0$

Using the quadratic formula to solve for $a$: $a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(3)(-2)}}{2(3)} = \frac{3 \pm \sqrt{9 + 24}}{6} = \frac{3 \pm \sqrt{33}}{6}$

Thus, we have two roots: $a_1 = \frac{3 + \sqrt{33}}{6} \quad \text{and} \quad a_2 = \frac{3 - \sqrt{33}}{6}$

Now, we check if these values belong to the given domain $a \in (-\infty, 0) \cup (1, \infty)$: Since $5 < \sqrt{33} < 6$:

• For $a_1$: $a_1 = \frac{3 + \sqrt{33}}{6} > \frac{3 + 5}{6} = \frac{8}{6} = \frac{4}{3} > 1 \implies a_1 \in (1, \infty)$
• For $a_2$: $a_2 = \frac{3 - \sqrt{33}}{6} < \frac{3 - 5}{6} = -\frac{2}{6} = -\frac{1}{3} < 0 \implies a_2 \in (-\infty, 0)$

Since both $a_1$ and $a_2$ lie in the given domain:

1. Both roots $a_1$ and $a_2$ are irrational numbers (since $\sqrt{33}$ is irrational). Therefore, Option C is TRUE.
2. There are two distinct values of $a$ that satisfy the equation. Therefore, Option D is TRUE.

Correct Options: (C) An irrational number $a$ satisfies the above equation (D) More than one $a$ satisfy the above equation