`L e t|tan^(-1)xtan^(-1)2xtan^(-1)3xtan^(-1)3xtan^(-1)xtan^(-1)2xtan^(-1)2xtan^(-1)3xtan^(-1)x|=0,t h e nt h enu m b e r` of values of `x` satisfying the equation is 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the given problem, we need to analyze the determinant:

\[ \left| \begin{array}{ccc} \tan^{-1} x & \tan^{-1} 2x & \tan^{-1} 3x \\ \tan^{-1} 3x & \tan^{-1} x & \tan^{-1} 2x \\ \tan^{-1} 2x & \tan^{-1} 3x & \tan^{-1} x \end{array} \right| = 0 \]

### Step 1: Define Variables
Let:
- \( a = \tan^{-1} x \)
- \( b = \tan^{-1} 2x \)
- \( c = \tan^{-1} 3x \)

Now, we can rewrite the determinant in terms of \( a \), \( b \), and \( c \):

\[ \left| \begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array} \right| = 0 \]

### Step 2: Use Determinant Properties
We can use the property of determinants that states if we add a multiple of one row to another, the value of the determinant remains unchanged. We will add the first row to the second and third rows:

\[ \left| \begin{array}{ccc} a & b & c \\ a+c & a+b & b+c \\ b+c & c+a & a+b \end{array} \right| = 0 \]

### Step 3: Factor Out Common Terms
Next, we can factor out \( a + b + c \) from the determinant:

\[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{array} \right| = 0 \]

### Step 4: Calculate the Determinant
Now, we calculate the determinant:

\[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{array} \right| = 1 \cdot (a^2 - bc) - 1 \cdot (b^2 - ac) + 1 \cdot (c^2 - ab) \]

This simplifies to:

\[ a^2 + b^2 + c^2 - ab - ac - bc = 0 \]

### Step 5: Analyze the Equation
The equation \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \) can be factored as:

\[ \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) = 0 \]

This implies that:

\[ a = b = c \]

### Step 6: Find Values of \( x \)
From \( a = \tan^{-1} x \), \( b = \tan^{-1} 2x \), and \( c = \tan^{-1} 3x \), we have:

\[ \tan^{-1} x = \tan^{-1} 2x = \tan^{-1} 3x \]

This leads to the equations:

1. \( x = 2x \) which gives \( x = 0 \)
2. \( x = 3x \) which gives \( x = 0 \)

Both equations yield \( x = 0 \) as the only solution.

### Conclusion
The number of values of \( x \) satisfying the equation is **1**.

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