If `tan^(-1).(1)/(2x+1)+tan^(-1).(1)/(4x+1)=cot^(-1)((x^(2))/(2))`, then the number of all possible values of x is/are

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the equation

\[ \tan^{-1}\left(\frac{1}{2x+1}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \cot^{-1}\left(\frac{x^2}{2}\right), \]

we can use the identity for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \]

provided \(ab < 1\).

### Step 1: Apply the identity

Let \(a = \frac{1}{2x+1}\) and \(b = \frac{1}{4x+1}\). Then we can write:

\[ \tan^{-1}\left(\frac{1}{2x+1}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{2x+1} \cdot \frac{1}{4x+1}}\right). \]

### Step 2: Simplify the right-hand side

Calculating \(a + b\):

\[ \frac{1}{2x+1} + \frac{1}{4x+1} = \frac{(4x+1) + (2x+1)}{(2x+1)(4x+1)} = \frac{6x + 2}{(2x+1)(4x+1)}. \]

Calculating \(1 - ab\):

\[ 1 - \frac{1}{(2x+1)(4x+1)} = \frac{(2x+1)(4x+1) - 1}{(2x+1)(4x+1)} = \frac{8x^2 + 6x}{(2x+1)(4x+1)}. \]

### Step 3: Combine the results

Now, substituting back, we have:

\[ \tan^{-1}\left(\frac{\frac{6x + 2}{(2x+1)(4x+1)}}{\frac{8x^2 + 6x}{(2x+1)(4x+1)}}\right) = \tan^{-1}\left(\frac{6x + 2}{8x^2 + 6x}\right). \]

### Step 4: Set the equation equal to cotangent

We know that:

\[ \cot^{-1}\left(\frac{x^2}{2}\right) = \tan^{-1}\left(\frac{2}{x^2}\right). \]

Thus, we equate:

\[ \frac{6x + 2}{8x^2 + 6x} = \frac{2}{x^2}. \]

### Step 5: Cross-multiply

Cross-multiplying gives:

\[ (6x + 2)x^2 = 2(8x^2 + 6x). \]

Expanding both sides:

\[ 6x^3 + 2x^2 = 16x^2 + 12x. \]

### Step 6: Rearranging the equation

Rearranging gives:

\[ 6x^3 - 14x^2 - 12x = 0. \]

### Step 7: Factor out common terms

Factoring out \(2x\):

\[ 2x(3x^2 - 7x - 6) = 0. \]

### Step 8: Solve the quadratic equation

Now we solve \(3x^2 - 7x - 6 = 0\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3}. \]

Calculating the discriminant:

\[ 49 + 72 = 121. \]

So we have:

\[ x = \frac{7 \pm 11}{6}. \]

Calculating the roots:

1. \(x = \frac{18}{6} = 3\)
2. \(x = \frac{-4}{6} = -\frac{2}{3}\)

### Step 9: Include the factor \(2x = 0\)

From \(2x = 0\), we have \(x = 0\).

### Final Step: Count all possible values

Thus, the possible values of \(x\) are \(3\), \(-\frac{2}{3}\), and \(0\). Therefore, the total number of possible values of \(x\) is **3**.