The equation `(x-3)^9+(x-3^2)^9+(x-3^3)^9+.....+(x-3^9)^9=0` has

Options

A. all real roots
B. real roots namely `3, 3^(2),……3^(9)`
C. one real & imaginary roots
D. none of these

Answer

To solve the equation \((x-3)^9 + (x-3^2)^9 + (x-3^3)^9 + \ldots + (x-3^9)^9 = 0\), we will analyze the function step by step.

### Step 1: Define the function
Let us define the function:
\[ f(x) = (x-3)^9 + (x-3^2)^9 + (x-3^3)^9 + \ldots + (x-3^9)^9 \]

### Step 2: Analyze the derivative
Next, we will find the derivative \(f'(x)\):
\[ f'(x) = 9(x-3)^8 + 9(x-3^2)^8 + 9(x-3^3)^8 + \ldots + 9(x-3^9)^8 \]
Factoring out the common term:
\[ f'(x) = 9\left((x-3)^8 + (x-3^2)^8 + (x-3^3)^8 + \ldots + (x-3^9)^8\right) \]

### Step 3: Determine the sign of the derivative
Each term \((x-3^k)^8\) for \(k = 1, 2, \ldots, 9\) is a non-negative quantity since it is raised to an even power. Therefore, \(f'(x) \geq 0\) for all \(x\), which means that \(f(x)\) is a non-decreasing function.

### Step 4: Evaluate the limits of the function
Now, let's evaluate the limits of \(f(x)\) as \(x\) approaches negative and positive infinity:
- As \(x \to -\infty\), each term \((x-3^k)^9\) tends to \(-\infty\), so \(f(x) \to -\infty\).
- As \(x \to +\infty\), each term \((x-3^k)^9\) tends to \(+\infty\), so \(f(x) \to +\infty\).

### Step 5: Apply the Intermediate Value Theorem
Since \(f(x)\) is continuous and goes from \(-\infty\) to \(+\infty\), by the Intermediate Value Theorem, there exists at least one real root \(c\) such that \(f(c) = 0\).

### Step 6: Conclude about the roots
Since \(f(x)\) is non-decreasing and crosses the x-axis only once, it implies that there is exactly one real root. The remaining roots must be imaginary.

### Final Answer
Thus, the equation has **one real root and the rest are imaginary roots**.

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