If the domain of the function

cos^-1((2x − 5)/(11x − 7)) + sin^-1(2x² − 3x + 1)

is [0, a] ∪ [12/13, b] then 1/(ab) is equal to

To find the domain of the function:
$f(x)\; =\; \backslash cos^\{-1\}\backslash left(\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\backslash right)\; +\; \backslash sin^\{-1\}(2x^2\; -\; 3x\; +\; 1)$
we need to find the values of $x$ for which both terms are defined simultaneously.

Step 1: Domain of the second term $\backslash sin^\{-1\}(2x^2\; -\; 3x\; +\; 1)$
For the arcsin function to be defined, its argument must lie in the interval $[-1,\; 1]$:
$-1\; \backslash le\; 2x^2\; -\; 3x\; +\; 1\; \backslash le\; 1$
This gives us two inequalities to solve:

First inequality:
$2x^2\; -\; 3x\; +\; 1\; \backslash le\; 1$
$2x^2\; -\; 3x\; \backslash le\; 0$
$x(2x\; -\; 3)\; \backslash le\; 0$
Which yields:
$x\; \backslash in\; \backslash left[0,\; \backslash frac\{3\}\{2\}\backslash right]$

Second inequality:
$2x^2\; -\; 3x\; +\; 1\; \backslash ge\; -1$
$2x^2\; -\; 3x\; +\; 2\; \backslash ge\; 0$
For this quadratic, the discriminant is . Since the coefficient of $x^2$ is positive and the discriminant is negative, this expression is strictly positive for all real numbers $x$.
Thus, the domain of the second term is:
$D\_2\; =\; \backslash left[0,\; \backslash frac\{3\}\{2\}\backslash right]$

Step 2: Domain of the first term $\backslash cos^\{-1\}\backslash left(\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\backslash right)$
For the arccos function to be defined, we must have:
$-1\; \backslash le\; \backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\; \backslash le\; 1$
This also gives us two inequalities:

First inequality:
$\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\; \backslash ge\; -1$
$\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\; +\; 1\; \backslash ge\; 0$
$\backslash frac\{2x\; -\; 5\; +\; 11x\; -\; 7\}\{11x\; -\; 7\}\; \backslash ge\; 0$
$\backslash frac\{13x\; -\; 12\}\{11x\; -\; 7\}\; \backslash ge\; 0$
The critical points are $x\; =\; \backslash frac\{7\}\{11\}$ and $x\; =\; \backslash frac\{12\}\{13\}$. Since $\backslash frac\{7\}\{11\}\; \backslash approx\; 0.636$ and $\backslash frac\{12\}\{13\}\; \backslash approx\; 0.923$, we have . Solving the rational inequality:
$x\; \backslash in\; \backslash left(-\backslash infty,\; \backslash frac\{7\}\{11\}\backslash right)\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; \backslash infty\backslash right)$

Second inequality:
$\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\; \backslash le\; 1$
$\backslash frac\{2x\; -\; 5\}\{11x\; -\; 7\}\; -\; 1\; \backslash le\; 0$
$\backslash frac\{2x\; -\; 5\; -\; (11x\; -\; 7)\}\{11x\; -\; 7\}\; \backslash le\; 0$
$\backslash frac\{-9x\; +\; 2\}\{11x\; -\; 7\}\; \backslash le\; 0$
$\backslash frac\{9x\; -\; 2\}\{11x\; -\; 7\}\; \backslash ge\; 0$
The critical points are $x\; =\; \backslash frac\{2\}\{9\}$ and $x\; =\; \backslash frac\{7\}\{11\}$. Since $\backslash frac\{2\}\{9\}\; \backslash approx\; 0.222$, we have . Solving this inequality:
$x\; \backslash in\; \backslash left(-\backslash infty,\; \backslash frac\{2\}\{9\}\backslash right]\; \backslash cup\; \backslash left(\backslash frac\{7\}\{11\},\; \backslash infty\backslash right)$

Intersecting the two cases for the first term:
$D\_1\; =\; \backslash left(\backslash left(-\backslash infty,\; \backslash frac\{7\}\{11\}\backslash right)\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; \backslash infty\backslash right)\backslash right)\; \backslash cap\; \backslash left(\backslash left(-\backslash infty,\; \backslash frac\{2\}\{9\}\backslash right]\; \backslash cup\; \backslash left(\backslash frac\{7\}\{11\},\; \backslash infty\backslash right)\backslash right)$
$D\_1\; =\; \backslash left(-\backslash infty,\; \backslash frac\{2\}\{9\}\backslash right]\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; \backslash infty\backslash right)$

Step 3: Intersection of the two domains
The domain of the overall function is the intersection of $D\_1$ and $D\_2$:
$Domain\; =\; D\_1\; \backslash cap\; D\_2\; =\; \backslash left(\backslash left(-\backslash infty,\; \backslash frac\{2\}\{9\}\backslash right]\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; \backslash infty\backslash right)\backslash right)\; \backslash cap\; \backslash left[0,\; \backslash frac\{3\}\{2\}\backslash right]$
Since , the intersection is:
$Domain\; =\; \backslash left[0,\; \backslash frac\{2\}\{9\}\backslash right]\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; \backslash frac\{3\}\{2\}\backslash right]$

Comparing this with the given domain $[0,\; a]\; \backslash cup\; \backslash left[\backslash frac\{12\}\{13\},\; b\backslash right]$, we find:
$a\; =\; \backslash frac\{2\}\{9\}$
$b\; =\; \backslash frac\{3\}\{2\}$

Now, we calculate the required expression:
$ab\; =\; \backslash frac\{2\}\{9\}\; \backslash times\; \backslash frac\{3\}\{2\}\; =\; \backslash frac\{1\}\{3\}$
Therefore:
$\backslash frac\{1\}\{ab\}\; =\; 3$