If there are only two linear functions f and g which map `[1,2] on [4,6]` and in a `DeltaABC, c =f (1)+g (1)` and a is the maximum valur of `r^(2),` where r is the distance of a variable point on the curve `x^(2)+y^(2)-xy=10` from the origin, then sin A: sin C is

Options

A. A. `1:2`
B. B. `2:1`
C. C. `1:1`
D. D. None of these

Answer

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Define the Linear Functions
We are given two linear functions \( f \) and \( g \) that map the interval \([1, 2]\) to \([4, 6]\). We can express these functions in the form:
\[ f(x) = ax + b \]
\[ g(x) = cx + d \]

### Step 2: Set Up the Equations for \( f \)
From the mapping:
- \( f(1) = 4 \) leads to the equation:
\[ a(1) + b = 4 \quad \text{(1)} \]
- \( f(2) = 6 \) leads to the equation:
\[ a(2) + b = 6 \quad \text{(2)} \]

### Step 3: Solve for \( a \) and \( b \)
Subtract equation (1) from equation (2):
\[ (2a + b) - (a + b) = 6 - 4 \]
This simplifies to:
\[ a = 2 \]
Now substitute \( a \) back into equation (1):
\[ 2 + b = 4 \implies b = 2 \]
Thus, we have:
\[ f(x) = 2x + 2 \]

### Step 4: Set Up the Equations for \( g \)
Now, for the function \( g \) which maps \([1, 2]\) to \([6, 4]\):
- \( g(1) = 6 \) leads to the equation:
\[ c(1) + d = 6 \quad \text{(3)} \]
- \( g(2) = 4 \) leads to the equation:
\[ c(2) + d = 4 \quad \text{(4)} \]

### Step 5: Solve for \( c \) and \( d \)
Subtract equation (3) from equation (4):
\[ (2c + d) - (c + d) = 4 - 6 \]
This simplifies to:
\[ c = -2 \]
Now substitute \( c \) back into equation (3):
\[ -2 + d = 6 \implies d = 8 \]
Thus, we have:
\[ g(x) = -2x + 8 \]

### Step 6: Calculate \( c \)
Now, we need to find \( c \) in the triangle \( ABC \):
\[ c = f(1) + g(1) = 4 + 6 = 10 \]

### Step 7: Find the Maximum Value of \( r^2 \)
We need to find the maximum value of \( r^2 \) where \( r \) is the distance of a variable point on the curve \( x^2 + y^2 - xy = 10 \).

Rearranging the equation:
\[ x^2 - xy + y^2 = 10 \]
This represents a conic section. To find the maximum distance from the origin, we can analyze the shape of this curve.

### Step 8: Convert to Standard Form
We can rewrite the equation as:
\[ \frac{(x - \frac{y}{2})^2}{10} + \frac{y^2}{\frac{40}{3}} = 1 \]
This indicates that it is an ellipse centered at the origin.

### Step 9: Identify the Major Axis
The maximum distance from the origin is given by the semi-major axis. The semi-major axis length is:
\[ \sqrt{10} \]
Thus, the maximum value of \( r^2 \) is:
\[ r^2 = 10 \]

### Step 10: Use the Sine Rule
We know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Given \( a = 10 \) and \( c = 10 \):
\[ \frac{10}{\sin A} = \frac{10}{\sin C} \]
This implies:
\[ \sin A = \sin C \]
Thus, the ratio \( \sin A : \sin C = 1 : 1 \).

### Final Answer
The ratio \( \sin A : \sin C \) is \( 1 : 1 \).

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