Consider the equation

${\int }_1^e\frac{{\left({\log }_{e}x\right)}^{\frac{1}{2}}}{x{(a-{\left({\log }_{e}x\right)}^{\frac{3}{2}})}^2}dx$ = 1

, where $a\in (-\infty ,0)\cup (1,\infty )$.
Which of the following statements is/are TRUE ?

Correct Statements:
1. An irrational number a satisfies the above equation
2. More than one a satisfy the above equation.

Let us evaluate the given definite integral step-by-step and determine all possible real values of $a$ that satisfy the equation.

The given integral equation is:

${\int }_1^e\frac{{\left({\log }_{e}x\right)}^{\frac{1}{2}}}{x{(a-{\left({\log }_{e}x\right)}^{\frac{3}{2}})}^2}dx=1$

Step 1: Use Substitution Method
Let $t={\log }_{e}x$. Then, its derivative with respect to $x$ is:

$dt=\frac{1}{x}dx$

Now, change the limits of integration according to $t$:

- When $x=1$, $t={\log }_{e}1=0$.
- When $x=e$, $t={\log }_{e}e=1$.

Substituting $t$ and $dt$ into the integral gives:

${\int }_0^1\frac{t^{\frac{1}{2}}}{{(a-t^{\frac{3}{2}})}^2}dt=1$

Step 2: Perform a Second Substitution
Let $u=a-t^{\frac{3}{2}}$.
Differentiating both sides gives:

$du=-\frac{3}{2}{t}^{\frac{1}{2}}dt\Rightarrow t^{\frac{1}{2}}dt=-\frac{2}{3}du$

Now, update the limits of integration for $u$:

- When $t=0$, $u=a-0^{\frac{3}{2}}=a$.
- When $t=1$, $u=a-1^{\frac{3}{2}}=a-1$.

Substituting $u$ and $du$ into the integral:

${\int }_a^{a-1}\frac{-\frac{2}{3}du}{u^2}=1$

Reversing the limits of integration to absorb the negative sign:

$\frac{2}{3}{\int }_{a-1}^a\frac{1}{u^2}du=1$

Step 3: Integrate and Solve for $a$
The antiderivative of $\frac{1}{u^2}$ is $-\frac{1}{u}$. Evaluating this between the limits:

$\frac{2}{3}{[-\frac{1}{u}]}_{a-1}^a=1$

$\frac{2}{3}(-\frac{1}{a}-(-\frac{1}{a-1}))=1$

$\frac{2}{3}(\frac{1}{a-1}-\frac{1}{a})=1$

Combine the fractions inside the parentheses:

$\frac{2}{3}\left(\frac{a-(a-1)}{a(a-1)}\right)=1$

$\frac{2}{3}\left(\frac{1}{a^2-a}\right)=1$

Cross-multiplying gives:

$a^2-a=\frac{2}{3}$

$3a^2-3a-2=0$

Step 4: Analyze the Roots of the Quadratic Equation
Using the quadratic formula $a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$a=\frac{3\pm \sqrt{{(-3)}^2-4\left(3\right)(-2)}}{2\left(3\right)}=\frac{3\pm \sqrt{9+24}}{6}=\frac{3\pm \sqrt{33}}{6}$

Since $\sqrt{33}$ is irrational, both roots:

$a_1=\frac{3+\sqrt{33}}{6}$ and $a_2=\frac{3-\sqrt{33}}{6}$

are irrational numbers.

Step 5: Check Domain Constraints
The given domain is $a\in (-\infty ,0)\cup (1,\infty )$.

- Since $5<\sqrt{33}<6$:
- $a_1=\frac{3+\sqrt{33}}{6}>\frac{3+5}{6}=\frac{8}{6}>1$, so $a_1\in (1,\infty )$.
- $a_2=\frac{3-\sqrt{33}}{6}<\frac{3-5}{6}=-\frac{2}{6}<0$, so $a_2\in (-\infty ,0)$.

Both roots lie completely within the allowed domain. Thus, there are two distinct irrational values of $a$ that satisfy the equation.

Therefore, the true statements are:

1. An irrational number a satisfies the above equation
2. More than one a satisfy the above equation.