Let `f (x)` be a continuous function (define for all x) which satisfies `f ^(3) (x)-5 f ^(2) (x)+ 10f (x) -12 ge 0, f ^(2) (x) + 3 ge 0 and f ^(2) (x) -5f(x)+ 6 le 0`
The equation of tangent that can be drawn from `(2,0)` on the curve `y = x^(2) f (sin x)` is :

Options

A. `y=24 (x+2)`
B. ` y= 12 (x+2)`
C. `y=24(x-2)`
D. `y=12 (x-2)`

Answer

To solve the problem, we need to find the equation of the tangent line to the curve \( y = x^2 f(\sin x) \) at the point where it touches the line from the point \( (2, 0) \). We also need to analyze the function \( f(x) \) based on the inequalities provided.

### Step 1: Analyze the inequalities for \( f(x) \)

1. **First Inequality**:
\[ f^{(3)}(x) - 5f^{(2)}(x) + 10f(x) - 12 \geq 0 \]
Let's denote \( f(x) \) as \( t \). Then we rewrite the inequality as:
\[ t^3 - 5t^2 + 10t - 12 \geq 0 \]
We can find the roots of the polynomial to analyze the intervals.

2. **Second Inequality**:
\[ f^{(2)}(x) + 3 \geq 0 \]
This implies:
\[ f^{(2)}(x) \geq -3 \]

3. **Third Inequality**:
\[ f^{(2)}(x) - 5f(x) + 6 \leq 0 \]
Rewriting gives:
\[ f^{(2)}(x) \leq 5f(x) - 6 \]

### Step 2: Solve the inequalities

1. From the first inequality, we can find the roots of \( t^3 - 5t^2 + 10t - 12 = 0 \). By testing values, we find that \( t = 3 \) is a root. We can factor the polynomial:
\[ (t - 3)(t^2 - 2t + 4) \geq 0 \]
The quadratic has no real roots (discriminant \( < 0 \)), so it is always positive. Thus, the inequality holds for:
\[ t \geq 3 \]

2. The second inequality \( f^{(2)}(x) + 3 \geq 0 \) implies:
\[ f^{(2)}(x) \geq -3 \]
This does not restrict \( f(x) \) further.

3. The third inequality \( f^{(2)}(x) - 5f(x) + 6 \leq 0 \) can be analyzed similarly. The roots of \( t^2 - 5t + 6 = 0 \) are \( t = 2 \) and \( t = 3 \). Thus, we have:
\[ 2 \leq f(x) \leq 3 \]

### Step 3: Determine \( f(x) \)

From the inequalities, we conclude that:
\[ f(x) = 3 \]
is a valid solution since it satisfies all inequalities.

### Step 4: Find the tangent line to the curve \( y = x^2 f(\sin x) \)

Substituting \( f(x) = 3 \):
\[ y = 3x^2 \]

### Step 5: Find the derivative

The derivative of \( y \) is:
\[ \frac{dy}{dx} = 6x \]

### Step 6: Find the slope at \( x = 2 \)

At \( x = 2 \):
\[ \frac{dy}{dx} = 6 \cdot 2 = 12 \]

### Step 7: Write the equation of the tangent line

Using point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2, 0) \) and \( m = 12 \):
\[ y - 0 = 12(x - 2) \]
\[ y = 12x - 24 \]

### Final Answer

The equation of the tangent line is:
\[ \boxed{y = 12x - 24} \]