Cosider the equation `int_(0)^(x) (t^(2)-8t+13)dt= x sin (a//x)`
If x takes the values for which the equation has a solution, then the number of values of `a in [0, 100]` is

Options

A. `y in [-5, 7]`
B. `y in [-7, 5]`
C. `y in [5, 7]`
D. none of these

Explanation

To solve the equation

\[ \int_{0}^{x} (t^2 - 8t + 13) \, dt = x \sin\left(\frac{a}{x}\right), \]

we will follow these steps:

### Step 1: Compute the Left-Hand Side (LHS)

We need to evaluate the integral on the left-hand side:

\[ \int_{0}^{x} (t^2 - 8t + 13) \, dt. \]

Calculating the integral:

\[ \int (t^2 - 8t + 13) \, dt = \frac{t^3}{3} - 4t^2 + 13t. \]

Now, we evaluate this from \(0\) to \(x\):

\[ \left[\frac{x^3}{3} - 4x^2 + 13x\right] - \left[\frac{0^3}{3} - 4(0^2) + 13(0)\right] = \frac{x^3}{3} - 4x^2 + 13x. \]

Thus, the LHS becomes:

\[ \frac{x^3}{3} - 4x^2 + 13x. \]

### Step 2: Set Up the Equation

Now, we set the LHS equal to the right-hand side (RHS):

\[ \frac{x^3}{3} - 4x^2 + 13x = x \sin\left(\frac{a}{x}\right). \]

### Step 3: Simplify the Equation

We can divide both sides by \(x\) (assuming \(x \neq 0\)):

\[ \frac{x^2}{3} - 4x + 13 = \sin\left(\frac{a}{x}\right). \]

### Step 4: Analyze the Function

Let

\[ f(x) = \frac{x^2}{3} - 4x + 13. \]

This is a quadratic function that opens upwards. We can find its vertex to determine its minimum value. The vertex \(x\) coordinate is given by:

\[ x = -\frac{b}{2a} = -\frac{-4}{2 \cdot \frac{1}{3}} = 6. \]

Now, substituting \(x = 6\) into \(f(x)\):

\[ f(6) = \frac{6^2}{3} - 4(6) + 13 = \frac{36}{3} - 24 + 13 = 12 - 24 + 13 = 1. \]

### Step 5: Determine the Range of \(f(x)\)

Since \(f(x)\) is a quadratic function that opens upwards, its minimum value is \(1\) at \(x = 6\). As \(x\) approaches \(0\) or increases indefinitely, \(f(x)\) will also increase indefinitely.

### Step 6: Set the Condition for \(\sin\left(\frac{a}{x}\right)\)

Since \(\sin\left(\frac{a}{x}\right)\) can only take values in the range \([-1, 1]\), we set the condition:

\[ f(x) \leq 1. \]

This leads to:

\[ \frac{x^2}{3} - 4x + 12 \leq 0. \]

### Step 7: Solve the Quadratic Inequality

To solve the quadratic inequality:

\[ x^2 - 12x + 36 \leq 0, \]

we can factor it:

\[ (x - 6)^2 \leq 0. \]

This implies:

\[ x = 6. \]

### Step 8: Find Values of \(a\)

Substituting \(x = 6\) into the equation:

\[ \sin\left(\frac{a}{6}\right) = 1. \]

This occurs when:

\[ \frac{a}{6} = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z}. \]

Thus,

\[ a = 3\pi + 12n\pi. \]

### Step 9: Determine the Values of \(a\) in the Range \([0, 100]\)

We need to find integer values of \(n\) such that:

\[ 0 \leq 3\pi + 12n\pi \leq 100. \]

Calculating the bounds:

\[ 3\pi \approx 9.42 \quad \Rightarrow \quad 12n\pi \leq 100 - 9.42 \quad \Rightarrow \quad 12n\pi \leq 90.58 \quad \Rightarrow \quad n \leq \frac{90.58}{12\pi} \approx 2.4. \]

Thus, \(n\) can take values \(0, 1, 2\).

Calculating the corresponding values of \(a\):

- For \(n = 0\): \(a = 3\pi \approx 9.42\)
- For \(n = 1\): \(a = 15\pi \approx 47.12\)
- For \(n = 2\): \(a = 27\pi \approx 84.82\)

### Conclusion

Thus, there are **3 values of \(a\)** in the range \([0, 100]\).