Statement-1: If a, b, c are distinct real numbers, then `a((x-b)(x-c))/((a-b)(a-c))+b((x-c)(x-a))/((b-c)(b-a))+c((x-a)(x-b))/((c-a)(c-b))=x` for each real x.
Statement-2: `If a, b, c in R` such that `ax^(2) + bx + c = 0` for three distinct real values of x, then `a = b = c = 0` i.e. `ax^(2) + bx + c = 0` for all `x in R`.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze the two statements provided and verify their validity step by step.

### Step 1: Understanding Statement 1
We need to show that the expression:
\[ f(x) = \frac{a(x-b)(x-c)}{(a-b)(a-c)} + \frac{b(x-c)(x-a)}{(b-c)(b-a)} + \frac{c(x-a)(x-b)}{(c-a)(c-b)} = x \]
holds true for all real \( x \) when \( a, b, c \) are distinct real numbers.

### Step 2: Evaluating \( f(a) \), \( f(b) \), and \( f(c) \)
Let’s evaluate \( f(x) \) at \( x = a \):
\[ f(a) = \frac{a(a-b)(a-c)}{(a-b)(a-c)} + \frac{b(a-c)(a-a)}{(b-c)(b-a)} + \frac{c(a-a)(a-b)}{(c-a)(c-b)} \]
The second and third terms become zero because they contain \( (a-a) \). Thus:
\[ f(a) = a \]

Now, evaluating \( f(b) \):
\[ f(b) = \frac{a(b-b)(b-c)}{(a-b)(a-c)} + \frac{b(b-c)(b-a)}{(b-c)(b-a)} + \frac{c(b-a)(b-b)}{(c-a)(c-b)} \]
Again, the first and third terms become zero. Thus:
\[ f(b) = b \]

Finally, evaluating \( f(c) \):
\[ f(c) = \frac{a(c-b)(c-c)}{(a-b)(a-c)} + \frac{b(c-c)(c-a)}{(b-c)(b-a)} + \frac{c(c-a)(c-b)}{(c-a)(c-b)} \]
Here, the first and second terms become zero. Thus:
\[ f(c) = c \]

### Step 3: Conclusion from Evaluations
Since \( f(a) = a \), \( f(b) = b \), and \( f(c) = c \), we can conclude that \( f(x) \) must be a polynomial of degree at most 2 (since it is defined by three distinct points). Given that it equals \( x \) at three distinct points, we can conclude that:
\[ f(x) = x \]
for all \( x \).

### Step 4: Understanding Statement 2
Statement 2 claims that if \( ax^2 + bx + c = 0 \) has three distinct real roots, then \( a = b = c = 0 \). This is true because a quadratic polynomial can have at most two distinct real roots. If it has three distinct roots, it must be the case that the polynomial is identically zero, which implies \( a = b = c = 0 \).

### Final Conclusion
Both statements are true, and Statement 2 correctly explains Statement 1.

### Final Answer
Thus, the answer is:
- Statement 1 is true.
- Statement 2 is a correct explanation of Statement 1.