निम्न समीकरण को हल कीजिए -
`tan^(-1)(x-1)+tan^(-1)(x)+tan^(-1)(x+1)=tan^(-1) 3 x `

Options

A. `-1`
B. `(1)/(2) `
C. 1
D. 2

Answer

To solve the equation \( \tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}(3x) \), we can use the properties of the inverse tangent function.

### Step 1: Combine the first two terms using the identity
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
for \( ab < 1 \).

Let \( a = x-1 \) and \( b = x \). Then,
\[ \tan^{-1}(x-1) + \tan^{-1}(x) = \tan^{-1}\left(\frac{(x-1) + x}{1 - (x-1)x}\right) \]
Calculating the numerator:
\[ (x-1) + x = 2x - 1 \]
Calculating the denominator:
\[ 1 - (x-1)x = 1 - (x^2 - x) = 1 + x - x^2 \]
Thus,
\[ \tan^{-1}(x-1) + \tan^{-1}(x) = \tan^{-1}\left(\frac{2x - 1}{1 + x - x^2}\right) \]

### Step 2: Combine the result with \( \tan^{-1}(x+1) \)
Now we need to add \( \tan^{-1}(x+1) \):
Let \( c = x + 1 \). So we have:
\[ \tan^{-1}\left(\frac{2x - 1}{1 + x - x^2}\right) + \tan^{-1}(x + 1) \]
Using the identity again:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
where \( a = \frac{2x - 1}{1 + x - x^2} \) and \( b = x + 1 \).

Calculating the numerator:
\[ \frac{2x - 1}{1 + x - x^2} + (x + 1) = \frac{(2x - 1) + (x + 1)(1 + x - x^2)}{1 + x - x^2} \]
Calculating the denominator:
\[ 1 - \left(\frac{2x - 1}{1 + x - x^2}\right)(x + 1) \]

### Step 3: Set the equation equal to \( \tan^{-1}(3x) \)
Now we have:
\[ \tan^{-1}\left(\text{Expression}\right) = \tan^{-1}(3x) \]
This implies:
\[ \text{Expression} = 3x \]

### Step 4: Solve for \( x \)
Equating the expressions from the previous steps and simplifying will lead to a polynomial equation in \( x \).

### Step 5: Find the roots
After simplifying, we arrive at:
\[ 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \]
Additionally, we also have \( x = 0 \) as a solution.

### Final Solutions
Thus, the solutions to the equation are:
\[ x = 0, \quad x = \frac{1}{2}, \quad x = -\frac{1}{2} \]