Let `f (x)=(x+1) (x+2) (x+3)…..(x+100) and g (x) =f (x) f''(x) -f'(x) ^(2).` Let n be the numbers of real roots of `g(x) =0,` then:

Options

A. `n lt 2`
B. `n gt 2`
C. `n lt 100`
D. `n gt 100`

Answer

To solve the problem, we need to analyze the function \( g(x) = f(x) f''(x) - (f'(x))^2 \) where \( f(x) = (x+1)(x+2)(x+3)\cdots(x+100) \).

### Step 1: Find \( f'(x) \)

We start by differentiating \( f(x) \). To do this, we can use logarithmic differentiation:

1. Take the natural logarithm of both sides:
\[ \ln f(x) = \ln((x+1)(x+2)(x+3)\cdots(x+100)) \]
This can be simplified using properties of logarithms:
\[ \ln f(x) = \ln(x+1) + \ln(x+2) + \cdots + \ln(x+100) \]

2. Differentiate both sides with respect to \( x \):
\[ \frac{f'(x)}{f(x)} = \frac{1}{x+1} + \frac{1}{x+2} + \cdots + \frac{1}{x+100} \]
Thus, we have:
\[ f'(x) = f(x) \left( \frac{1}{x+1} + \frac{1}{x+2} + \cdots + \frac{1}{x+100} \right) \]

### Step 2: Find \( f''(x) \)

Next, we differentiate \( f'(x) \) to find \( f''(x) \):
1. We apply the product rule:
\[ f''(x) = \left( f'(x) \right)' = \left( f(x) \left( \frac{1}{x+1} + \frac{1}{x+2} + \cdots + \frac{1}{x+100} \right) \right)' \]
This requires applying the product rule and the quotient rule.

2. After differentiating, we can express \( f''(x) \) in terms of \( f(x) \) and \( f'(x) \).

### Step 3: Analyze \( g(x) \)

Now we substitute \( f(x) \), \( f'(x) \), and \( f''(x) \) into \( g(x) \):
\[ g(x) = f(x) f''(x) - (f'(x))^2 \]

### Step 4: Set \( g(x) = 0 \)

To find the roots of \( g(x) \), we set:
\[ f(x) f''(x) - (f'(x))^2 = 0 \]

This can be rearranged to:
\[ f(x) f''(x) = (f'(x))^2 \]

### Step 5: Analyze the roots

1. We know that \( f(x) \) is a polynomial of degree 100, which means it has 100 roots (real and complex).
2. The expression \( f(x) f''(x) - (f'(x))^2 \) is a result of the Wronskian determinant of \( f \) and its derivatives.

### Step 6: Determine the number of real roots

1. Since \( f(x) \) has 100 roots and \( f'(x) \) has 99 roots, we can analyze the behavior of \( g(x) \).
2. The function \( g(x) \) is a polynomial and can change signs at the roots of \( f(x) \) and \( f'(x) \).
3. However, since \( f(x) \) is always positive between its roots and \( f'(x) \) is also positive, \( g(x) \) does not cross zero.

### Conclusion

Thus, the number of real roots \( n \) of \( g(x) = 0 \) is:
\[ n = 0 \]

### Final Answer
The number of real roots \( n \) is \( 0 \).

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