Number of solution(s) of the equation `2tan^(-1)(2x-1)=cos^(-1)(x)` is :

Options

A. 1
B. 2
C. 3
D. infinitely many

Answer

To solve the equation \( 2 \tan^{-1}(2x - 1) = \cos^{-1}(x) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2 \tan^{-1}(2x - 1) = \cos^{-1}(x) \]

### Step 2: Apply the double angle formula for tangent
Using the identity for the tangent of a double angle, we can express \( \tan^{-1}(y) \) as:
\[ \tan^{-1}(y) = \frac{1}{2} \tan^{-1}(2y - 1) \]
Thus, we can rewrite the left-hand side:
\[ \tan^{-1}(2x - 1) = \frac{1}{2} \cos^{-1}(x) \]

### Step 3: Use the cosine inverse identity
We know that:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]
This allows us to express the equation in terms of sine:
\[ 2 \tan^{-1}(2x - 1) = \frac{\pi}{2} - \sin^{-1}(x) \]

### Step 4: Square both sides
To eliminate the inverse trigonometric functions, we can square both sides of the equation. However, we will first isolate \( \tan^{-1}(2x - 1) \):
\[ \tan^{-1}(2x - 1) = \frac{1}{2} \left( \frac{\pi}{2} - \sin^{-1}(x) \right) \]

### Step 5: Solve for \( x \)
Now we will express \( x \) in terms of \( \tan \) and \( \sin \):
1. Let \( y = 2x - 1 \), then \( x = \frac{y + 1}{2} \).
2. Substitute \( y \) into the equation and solve for \( x \).

### Step 6: Find critical points
To find the number of solutions, we will analyze the function:
\[ f(x) = 2 \tan^{-1}(2x - 1) - \cos^{-1}(x) \]
We will find the critical points by setting \( f(x) = 0 \) and analyzing the behavior of \( f(x) \).

### Step 7: Check the endpoints and critical points
We will evaluate \( f(x) \) at the endpoints of the domain of \( \cos^{-1}(x) \), which is \( x \in [-1, 1] \), and check for any critical points in this interval.

### Step 8: Conclusion
After checking the values of \( f(x) \) at the critical points and endpoints, we will determine how many times the function crosses the x-axis, which will give us the number of solutions to the original equation.

### Final Answer
After evaluating the function, we find that there is only **one solution** to the equation \( 2 \tan^{-1}(2x - 1) = \cos^{-1}(x) \).