Question

Let \( S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1} (2x + 1) \ \). Then \[ \sum_{x \in S} (2x - 1)^2 \text{ is equal to:} \]

Answer 1

Correct Answer: 5

Given the equation: $\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$

We can rewrite this as:

$2 \cos^{-1} x - \sin^{-1}(2x + 1) = \frac{3\pi}{2}$

1. Substitution:
Let $\cos^{-1} x = \alpha$ and $\sin^{-1}(2x + 1) = \beta$. Then the equation becomes:

$2\alpha - \beta = \frac{3\pi}{2}$

Rearranging the terms, we get:

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

2. Taking the Cosine of Both Sides:
Taking the cosine of both sides:

$\cos 2\alpha = \cos \left( \frac{3\pi}{2} + \beta \right)$

Using the identity $\cos \left( \frac{3\pi}{2} + \beta \right) = \sin \beta$, we have:

$\cos 2\alpha = \sin \beta$

3. Using Trigonometric Identities:
Using the double angle formula $\cos 2\alpha = 2 \cos^2 \alpha - 1$, and knowing that $\cos \alpha = x$ and $\sin \beta = 2x + 1$, we substitute these values:

$2x^2 - 1 = 2x + 1$

4. Rearranging into a Quadratic Equation:
Rearranging the terms to form a quadratic equation:

$x^2 - x - 1 = 0$

5. Solving the Quadratic Equation:
Using the quadratic formula, we find the solutions for $x$:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

6. Validating the Solutions:
We need to check the validity of these solutions. The range of $\cos^{-1} x$ is $[0, \pi]$ and the range of $\sin^{-1}(2x+1)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

If $x = \frac{1 + \sqrt{5}}{2} \approx 1.618$, this is outside the domain of $\cos^{-1} x$, which is $[-1, 1]$. Therefore, this solution is rejected.

If $x = \frac{1 - \sqrt{5}}{2} \approx -0.618$, this is within the domain of $\cos^{-1} x$. Let's consider the expression we need to evaluate:

$4x^2 - 4x = 4(x^2 - x)$

From the quadratic equation $x^2 - x - 1 = 0$, we have $x^2 - x = 1$.

Therefore, $4x^2 - 4x = 4(1) = 4$.

7. Calculating $(2x - 1)^2$:
However, the question asks for the value of $(2x - 1)^2$. Let's calculate that:

$2x - 1 = 2 \left( \frac{1 - \sqrt{5}}{2} \right) - 1 = 1 - \sqrt{5} - 1 = -\sqrt{5}$

$(2x - 1)^2 = (-\sqrt{5})^2 = 5$

Final Answer:
The value of $(2x - 1)^2$ is $ \boxed{5} $.