Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f(x)={\cos }^{-1}(\frac{2x-5}{11-3x})+{\sin }^{-1}(2x^2-3x+1)$ is the interval $[\alpha ,\beta ]$, then $\alpha +2\beta$ is equal to :

Unlocking the Domain

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey to define the very boundaries of existence for a function. In the world of JEE Advanced, the 'domain' is the most fundamental concept.

If you cannot find the domain, you cannot integrate, you cannot differentiate, and you cannot analyze the behavior of the function. Let us master this.

We are presented with the function $f(x)={\cos }^{-1}(\frac{2x-5}{11-3x})+{\sin }^{-1}(2x^2-3x+1)$. This is a composite function, a sum of two distinct inverse trigonometric entities.

For $f(x)$ to be defined, both parts must exist simultaneously. This means we are looking for the intersection of two sets: the domain of the ${\cos }^{-1}$ term ($D_1$) and the domain of the ${\sin }^{-1}$ term ($D_2$). Our final answer will be $D=D_1\cap D_2$.

The Rational Challenge

Let us focus on the first term: ${\cos }^{-1}(\frac{2x-5}{11-3x})$. We know that the input of ${\cos }^{-1}(u)$ must lie in the closed interval $[-1,1]$. Therefore, we must satisfy the double inequality:

Here is where many students stumble. Do not, under any circumstances, cross-multiply by $(11-3x)$. Because we do not know if $(11-3x)$ is positive or negative, cross-multiplying could flip the inequality sign and cause you to lose your answer.

Instead, we treat this as two separate inequalities: $\frac{2x-5}{11-3x}\ge -1$ and $\frac{2x-5}{11-3x}\le 1$.

For the first part, $\frac{2x-5}{11-3x}+1\ge 0$, we find a common denominator:

Multiplying by $-1$ to make the leading coefficients positive gives us $\frac{x-6}{3x-11}\ge 0$. Using the Wavy Curve method, we identify the critical points at $x=\frac{11}{3}$ and $x=6$. Testing the intervals, we find the solution: $x\in (-\infty ,\frac{11}{3})\cup [6,\infty )$.

For the second part, $\frac{2x-5}{11-3x}-1\le 0$, we get:

Using the Wavy Curve method with critical points $x=\frac{11}{3}$ and $x=\frac{16}{5}$, we find the solution: $x\in (-\infty ,\frac{16}{5}]\cup (\frac{11}{3},\infty )$. Intersecting these two sets gives us our first domain:

The Quadratic Stability

Now, let us turn our attention to the second term: ${\sin }^{-1}(2x^2-3x+1)$. Again, the input must lie in $[-1,1]$. So, $-1\le 2x^2-3x+1\le 1$.

First, consider $2x^2-3x+1\ge -1$, which becomes $2x^2-3x+2\ge 0$. Let us check the discriminant:

Since the discriminant is negative and the leading coefficient is positive, this parabola never touches the $x$-axis and is always positive. This inequality is true for all real numbers $x\in \mathbb{R}$.

Next, consider $2x^2-3x+1\le 1$, which simplifies to $2x^2-3x\le 0$. Factoring this, we get $x(2x-3)\le 0$. The roots are $0$ and $\frac{3}{2}$. The inequality holds between these roots, so $x\in [0,\frac{3}{2}]$.

Intersecting $\mathbb{R}$ with $[0,\frac{3}{2}]$ gives us:

The Grand Convergence

We have arrived at the final step. We need the intersection of $D_1=(-\infty ,\frac{16}{5}]\cup [6,\infty )$ and $D_2=[0,\frac{3}{2}]$.

Visualize the number line. $D_2$ is the interval $[0,1.5]$. $D_1$ covers everything up to $3.2$ and everything from $6$ onwards.

Since $1.5<3.2$, the entire interval $[0,1.5]$ is contained within the first part of $D_1$. Therefore, the intersection is simply:

Comparing this to the given interval $[\alpha ,\beta ]$, we identify $\alpha =0$ and $\beta =\frac{3}{2}$. The problem asks for $\alpha +2\beta$. Substituting our values: