Statement -1: The equation who roots are reciprocal of the roots of the equation `10x^(2) -x-5=0" is " 5x^(2) +x-10=0`
and
Statement -2 : To obtain a quadratic equation whose roots are reciprocal of the roots of the given equation `ax^(2) +bx +c=0` change the coefficients a,b,c,to c,b,a. `(c ne 0)`

Options

A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -11
B. Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -11
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is Flase, Statement -2 is True

Answer

To solve the problem, we need to verify both statements regarding the quadratic equations and their roots.

### Given:
1. **Statement 1**: The equation whose roots are the reciprocals of the roots of \(10x^2 - x - 5 = 0\) is \(5x^2 + x - 10 = 0\).
2. **Statement 2**: To obtain a quadratic equation whose roots are the reciprocals of the roots of the given equation \(ax^2 + bx + c = 0\), change the coefficients \(a, b, c\) to \(c, b, a\) (where \(c \neq 0\)).

### Step-by-Step Solution:

#### Step 1: Find the roots of the equation \(10x^2 - x - 5 = 0\).
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 10\), \(b = -1\), and \(c = -5\).

Calculating the discriminant:
\[ D = b^2 - 4ac = (-1)^2 - 4 \cdot 10 \cdot (-5) = 1 + 200 = 201 \]

Now, substituting into the quadratic formula:
\[ x = \frac{1 \pm \sqrt{201}}{20} \]

Let the roots be:
\[ \alpha = \frac{1 + \sqrt{201}}{20}, \quad \beta = \frac{1 - \sqrt{201}}{20} \]

#### Step 2: Find the reciprocals of the roots.
The reciprocals of the roots are:
\[ \frac{1}{\alpha} = \frac{20}{1 + \sqrt{201}}, \quad \frac{1}{\beta} = \frac{20}{1 - \sqrt{201}} \]

#### Step 3: Form the quadratic equation with these roots.
The sum of the reciprocals:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\frac{1 - \sqrt{201}}{20} + \frac{1 + \sqrt{201}}{20}}{\alpha \beta} \]
Calculating \(\alpha \beta\):
\[ \alpha \beta = \frac{-5}{10} = -\frac{1}{2} \]
Thus,
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{1/20}{-\frac{1}{2}} = -\frac{1}{10} \]

The product of the reciprocals:
\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = -2 \]

Using the sum and product to form the quadratic equation:
\[ x^2 - \left(-\frac{1}{10}\right)x - 2 = 0 \implies 10x^2 + x - 20 = 0 \]

To match the form \(5x^2 + x - 10 = 0\), divide the entire equation by 2:
\[ 5x^2 + x - 10 = 0 \]

#### Conclusion for Statement 1:
Thus, **Statement 1 is correct**.

#### Step 4: Verify Statement 2.
To obtain a quadratic equation whose roots are the reciprocals of the roots of \(ax^2 + bx + c = 0\), we replace \(a, b, c\) with \(c, b, a\). This is indeed correct as shown in the derivation above.

### Conclusion for Statement 2:
Thus, **Statement 2 is also correct**.

### Final Answer:
Both statements are correct.