If the domain of the function f(x) = (cos⁻¹ √(x²-x+1))/(√(sin⁻¹((2x-1)/2))) is the interval (α, β],

Q: Wait, why do we reject the negative root for the numerator's $\cos^{-1}$ constraint?

Great observation! The domain of $\cos^{-1}(u)$ is indeed $[-1, 1]$. However, our argument $u$ is $\sqrt{x^2-x+1}$. Since the output of a square root function is by definition non-negative, $u$ can never be negative. Thus, the lower bound of the interval $[-1, 1]$ is automatically satisfied, and we only need to focus on the upper bound, $u \le 1$.

Q: Why is the denominator condition $\sin^{-1}(\dots) > 0$ and not $\ge 0$?

This is a classic JEE trap! The expression is $\sqrt{\sin^{-1}(\dots)}$. If the term inside the square root were zero, the denominator would become zero, making the function undefined. Therefore, we must exclude the case where the expression equals zero, leaving us with the strict inequality $\sin^{-1}(\dots) > 0$.

Q: Is there a trick with the quadratic $x^2-x+1$ in the numerator?

Yes, it's a subtle one. If you calculate the discriminant $D = b^2 - 4ac$, you get $1 - 4 = -3$. Since $D < 0$ and the coefficient of $x^2$ is positive, the quadratic $x^2-x+1$ is always positive for all real $x$. This simplifies our work significantly because the condition $x^2-x+1 \ge 0$ is always true, allowing us to focus entirely on the upper bound $x^2-x+1 \le 1$.

Question

If the domain of the function $f(x) = \frac{\cos^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin^{-1}(\frac{2x-1}{2})}}$ is the interval $(\alpha, \beta]$, then $\alpha + \beta$ is equal to :

Options:

(A) $\frac{3}{2}$

(B) 2

(C) $\frac{1}{2}$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a grand, complex mathematical structure. To enter, you must satisfy the rules of every gatekeeper guarding the path.

In our function

$$f(x) = \frac{\cos^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin^{-1}(\frac{2x-1}{2})}}$$

we have two primary gatekeepers: the numerator and the denominator. To find the domain, we must find the values of $x$ that satisfy both simultaneously.

### Phase 1: The Numerator's Constraint

The numerator is $\cos^{-1} \sqrt{x^2-x+1}$. We know that the inverse cosine function, $\cos^{-1}(u)$, is a picky gatekeeper; it only accepts inputs $u$ in the range $[-1, 1]$.

Here, our input is $u = \sqrt{x^2-x+1}$. So, we must have $-1 \le \sqrt{x^2-x+1} \le 1$.

Since a square root is always non-negative, the condition $-1 \le \sqrt{x^2-x+1}$ is always true for all real $x$. We are left with the condition:

$$\sqrt{x^2-x+1} \le 1$$

Squaring both sides, we get $x^2-x+1 \le 1$. Subtracting $1$ from both sides, we arrive at $x^2-x \le 0$, which factors into $x(x-1) \le 0$.

Using the wavy curve method, we see this inequality holds when $x \in [0, 1]$.

### Phase 2: The Denominator's Gate

Now, we face the denominator: $\sqrt{\sin^{-1}(\frac{2x-1}{2})}$. This gatekeeper is even stricter.

First, the expression inside the square root must be non-negative: $\sin^{-1}(\frac{2x-1}{2}) \ge 0$. Second, because it sits in the denominator, it cannot be zero.

Combining these, we need:

$$\sin^{-1}\left(\frac{2x-1}{2}\right) > 0$$

For the inverse sine function to be strictly positive, its argument must be strictly greater than $0$ and less than or equal to $1$. Thus, we set up the compound inequality:

$$0 < \frac{2x-1}{2} \le 1$$

Multiplying by $2$, we get $0 < 2x-1 \le 2$. Adding $1$ to all parts, we have $1 < 2x \le 3$.

Finally, dividing by $2$, we find $x \in (\frac{1}{2}, \frac{3}{2}]$.

### The Intersection and Final Calculation

We have two intervals: the numerator's domain $[0, 1]$ and the denominator's domain $(\frac{1}{2}, \frac{3}{2}]$. For the function to exist, $x$ must live in both worlds.

The intersection of $[0, 1]$ and $(\frac{1}{2}, \frac{3}{2}]$ is the interval $(\frac{1}{2}, 1]$.

Comparing this to the given form $(\alpha, \beta]$, we identify $\alpha = \frac{1}{2}$ and $\beta = 1$. The final step is simple arithmetic:

$$\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}$$

You have successfully navigated the constraints and unlocked the solution. The final answer is **$\frac{3}{2}$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f (x) = \frac{c o s ^{- 1} x ^{2} - x + 1}{s i n ^{- 1} ( \frac{2 x - 1}{2} )}$ is the interval $(α, β]$ , then $α + β$ is equal to :

Select Answer:

Visualized Solution

Function Decomposition

Numerator Constraint: $cos^{- 1}$

Squaring the Numerator Inequality

Solving for $x$ in Numerator

Denominator Constraint

Sine Inverse Argument Range

Solving for $x$ in Denominator

Finding the Intersection

Final Calculation

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Analyzing the Setup

Phase 1

Phase 2

The Intersection and Final Calculation

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Select Answer:

Visualized Solution

Function Decomposition

Numerator Constraint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos−1

Squaring the Numerator Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in Numerator

Denominator Constraint

Sine Inverse Argument Range

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in Denominator

Finding the Intersection

Final Calculation

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Analyzing the Setup

Phase 1

Phase 2

The Intersection and Final Calculation

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