If two curves `y=2 sin ((5pix)/6)` and `y=alphax^(2)-3alphax+2alpha+1` touch each other at some point then the value of `(sqrt(3)alpha)/(5pi) " is "(0le x le (18)/(5))`

Answer

To solve the problem step by step, we need to find the value of \(\frac{\sqrt{3}\alpha}{5\pi}\) given that the two curves \(y = 2 \sin\left(\frac{5\pi x}{6}\right)\) and \(y = \alpha x^2 - 3\alpha x + 2\alpha + 1\) touch each other at some point.

### Step 1: Set the equations equal to each other
Since the curves touch each other at some point, we can set their equations equal to each other:
\[ 2 \sin\left(\frac{5\pi x}{6}\right) = \alpha x^2 - 3\alpha x + 2\alpha + 1 \]
This gives us the equation:
\[ 2 \sin\left(\frac{5\pi x}{6}\right) - \alpha x^2 + 3\alpha x - (2\alpha + 1) = 0 \]

### Step 2: Find a point of intersection
To find a point where these curves touch, we can test some values of \(x\). Let's start with \(x = 1\):
\[ 2 \sin\left(\frac{5\pi \cdot 1}{6}\right) = 2 \sin\left(\frac{5\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1 \]
Now substituting \(x = 1\) into the second curve:
\[ \alpha(1)^2 - 3\alpha(1) + 2\alpha + 1 = \alpha - 3\alpha + 2\alpha + 1 = 1 \]
Thus, at \(x = 1\), both curves equal \(1\). Therefore, the point of intersection is \(P(1, 1)\).

### Step 3: Find the derivatives of both curves
Next, we need to find the derivatives of both functions to ensure they have the same slope at the point of intersection.

For the first curve:
\[ y_1 = 2 \sin\left(\frac{5\pi x}{6}\right) \]
The derivative is:
\[ y_1' = 2 \cdot \frac{5\pi}{6} \cos\left(\frac{5\pi x}{6}\right) = \frac{5\pi}{3} \cos\left(\frac{5\pi x}{6}\right) \]

For the second curve:
\[ y_2 = \alpha x^2 - 3\alpha x + 2\alpha + 1 \]
The derivative is:
\[ y_2' = 2\alpha x - 3\alpha \]

### Step 4: Evaluate the derivatives at \(x = 1\)
Now we evaluate both derivatives at \(x = 1\):
\[ y_1'(1) = \frac{5\pi}{3} \cos\left(\frac{5\pi \cdot 1}{6}\right) = \frac{5\pi}{3} \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{5\pi \sqrt{3}}{6} \]
\[ y_2'(1) = 2\alpha(1) - 3\alpha = -\alpha \]

### Step 5: Set the derivatives equal
Since the curves touch, their derivatives must also be equal at \(x = 1\):
\[ -\alpha = -\frac{5\pi \sqrt{3}}{6} \]
Thus, we have:
\[ \alpha = \frac{5\pi \sqrt{3}}{6} \]

### Step 6: Find \(\frac{\sqrt{3}\alpha}{5\pi}\)
Now we substitute \(\alpha\) into \(\frac{\sqrt{3}\alpha}{5\pi}\):
\[ \frac{\sqrt{3}\alpha}{5\pi} = \frac{\sqrt{3} \cdot \frac{5\pi \sqrt{3}}{6}}{5\pi} = \frac{5 \cdot 3}{6} = \frac{15}{6} = \frac{5}{2} \]

### Final Answer
Thus, the value of \(\frac{\sqrt{3}\alpha}{5\pi}\) is:
\[ \frac{5}{2} \]