p(x) be a polynomial of degree at most 5 which leaves remainder - 1 and 1 upon division by `(x-1)^3` and `(x+1)^3`respectively, the number of real roots of `P(x) = 0` is (a) 1 (b) 3 (c) 5 (d)2

Options

A. `1`
B. `3`
C. `5`
D. `2`

Explanation

To solve the problem, we need to determine the number of real roots of the polynomial \( P(x) \) given the conditions about its remainders when divided by \( (x-1)^3 \) and \( (x+1)^3 \).

### Step 1: Understanding the Polynomial
Since \( P(x) \) is a polynomial of degree at most 5, we can express it in the form:
\[ P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 \]

### Step 2: Applying the Remainder Theorem
According to the problem, when \( P(x) \) is divided by \( (x-1)^3 \), the remainder is -1. This means:
\[ P(1) = -1, \quad P'(1) = 0, \quad P''(1) = 0 \]
Similarly, when \( P(x) \) is divided by \( (x+1)^3 \), the remainder is 1:
\[ P(-1) = 1, \quad P'(-1) = 0, \quad P''(-1) = 0 \]

### Step 3: Setting Up the System of Equations
From the above conditions, we can set up the following equations:

1. \( P(1) = a_5 + a_4 + a_3 + a_2 + a_1 + a_0 = -1 \)
2. \( P' (1) = 5a_5 + 4a_4 + 3a_3 + 2a_2 + a_1 = 0 \)
3. \( P''(1) = 20a_5 + 12a_4 + 6a_3 + 2a_2 = 0 \)

4. \( P(-1) = -a_5 + a_4 - a_3 + a_2 - a_1 + a_0 = 1 \)
5. \( P' (-1) = -5a_5 + 4a_4 - 3a_3 + 2a_2 - a_1 = 0 \)
6. \( P'' (-1) = 20a_5 + 12a_4 + 6a_3 + 2a_2 = 0 \)

### Step 4: Analyzing the Roots
From the conditions \( P'(1) = 0 \) and \( P'(-1) = 0 \), we know that \( x = 1 \) and \( x = -1 \) are double roots of \( P(x) \).

### Step 5: Formulating the Polynomial
Since \( P(x) \) has double roots at \( x = 1 \) and \( x = -1 \), we can express \( P(x) \) as:
\[ P(x) = k(x-1)^2(x+1)^2(x-r) \]
where \( r \) is another root (which can be real or complex) and \( k \) is a constant.

### Step 6: Finding the Number of Real Roots
The polynomial \( P(x) \) has:
- Two double roots at \( x = 1 \) and \( x = -1 \).
- The term \( (x-r) \) can contribute either one more real root or none depending on the value of \( r \).

Thus, the total number of real roots can be:
- If \( r \) is real: 5 real roots (2 from \( x=1 \), 2 from \( x=-1 \), and 1 from \( r \)).
- If \( r \) is complex: 4 real roots (2 from \( x=1 \), 2 from \( x=-1 \)).

### Conclusion
Since \( r \) can be real or complex, the maximum number of real roots is 5. However, the problem asks for the number of real roots of \( P(x) = 0 \), which can be concluded as:

The number of real roots of \( P(x) = 0 \) is **3** (two double roots and one additional real root).

### Final Answer
Thus, the answer is:
(b) 3