Statement-1 `sum_(r=0)^(n) r ""^(n)C_(r) x^(r) (-1)^(r) = nx (1 - x)^(n -1)`
Statement-2:` sum_(r=0)^(n)r ""^(n)C_(r) x^(r) (-1)^(r) =0`

Options

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

Explanation

To solve the problem, we need to analyze both statements and determine their validity step by step.

### Step 1: Analyze Statement 2
We start with Statement 2:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^r (-1)^r = 0 \]
Using the identity for \( r \binom{n}{r} = n \binom{n-1}{r-1} \), we can rewrite the left-hand side:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^r (-1)^r = n \sum_{r=1}^{n} \binom{n-1}{r-1} x^r (-1)^r \]
Changing the index of summation by letting \( k = r - 1 \), we have:
\[ = n \sum_{k=0}^{n-1} \binom{n-1}{k} x^{k+1} (-1)^{k+1} \]
This can be simplified further:
\[ = -n x \sum_{k=0}^{n-1} \binom{n-1}{k} x^k (-1)^k \]
Using the Binomial Theorem, we know:
\[ \sum_{k=0}^{n-1} \binom{n-1}{k} (-x)^k = (1 - x)^{n-1} \]
Thus, we have:
\[ -n x (1 - x)^{n-1} \]
Setting this equal to zero gives us:
\[ -n x (1 - x)^{n-1} = 0 \]
This is true for \( x = 0 \) or \( x = 1 \), or for any \( n \) when \( x \) is in the range [0, 1]. Therefore, Statement 2 is **true**.

### Step 2: Analyze Statement 1
Now we check Statement 1:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^r (-1)^r = n x (1 - x)^{n-1} \]
From our previous calculation, we found that:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^r (-1)^r = -n x (1 - x)^{n-1} \]
This is not equal to \( n x (1 - x)^{n-1} \) since it has a negative sign. Therefore, Statement 1 is **false**.

### Conclusion
- Statement 1 is false.
- Statement 2 is true.

### Final Answer
Thus, the correct answer is that Statement 1 is false and Statement 2 is true.

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