The minimum integral value of `alpha` for which the quadratic equation `(cot^(-1)alpha)x^(2)-(tan^(-1)alpha)^(3//2)x+2(cot^(-1)alpha)^(2)=0` has both positive roots

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the minimum integral value of \( \alpha \) for which the quadratic equation

\[ (\cot^{-1} \alpha)x^2 - (\tan^{-1} \alpha)^{3/2}x + 2(\cot^{-1} \alpha)^2 = 0 \]

has both positive roots, we will follow these steps:

### Step 1: Identify the coefficients of the quadratic equation
The coefficients of the quadratic equation are:
- \( a = \cot^{-1} \alpha \)
- \( b = -(\tan^{-1} \alpha)^{3/2} \)
- \( c = 2(\cot^{-1} \alpha)^2 \)

### Step 2: Condition for real roots
For the quadratic equation to have real roots, the discriminant must be non-negative. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Substituting the coefficients:

\[ D = \left(-(\tan^{-1} \alpha)^{3/2}\right)^2 - 4(\cot^{-1} \alpha)(2(\cot^{-1} \alpha)^2) \]

This simplifies to:

\[ D = (\tan^{-1} \alpha)^{3} - 8(\cot^{-1} \alpha)^3 \]

Setting the discriminant \( D \geq 0 \):

\[ (\tan^{-1} \alpha)^{3} \geq 8(\cot^{-1} \alpha)^3 \]

### Step 3: Using the identity
Recall that:

\[ \tan^{-1} \alpha + \cot^{-1} \alpha = \frac{\pi}{2} \]

Thus, we can express \( \cot^{-1} \alpha \) in terms of \( \tan^{-1} \alpha \):

\[ \cot^{-1} \alpha = \frac{\pi}{2} - \tan^{-1} \alpha \]

### Step 4: Substitute back into the inequality
Substituting this into our inequality gives:

\[ (\tan^{-1} \alpha)^{3} \geq 8\left(\frac{\pi}{2} - \tan^{-1} \alpha\right)^{3} \]

### Step 5: Analyze the roots
For the quadratic to have both roots positive, the sum of the roots must be positive. The sum of the roots is given by:

\[ -\frac{b}{a} = \frac{(\tan^{-1} \alpha)^{3/2}}{\cot^{-1} \alpha} \]

This must be greater than zero, which implies:

\[ (\tan^{-1} \alpha)^{3/2} > 0 \quad \text{and} \quad \cot^{-1} \alpha > 0 \]

### Step 6: Determine conditions for \( \alpha \)
Both conditions imply that \( \alpha > 0 \). Additionally, we need the product of the roots to be positive, which is given by:

\[ \frac{c}{a} = \frac{2(\cot^{-1} \alpha)^2}{\cot^{-1} \alpha} = 2\cot^{-1} \alpha > 0 \]

This condition is satisfied when \( \alpha > 0 \).

### Step 7: Solve for \( \alpha \)
From the earlier step, we derived that:

\[ \tan^{-1} \alpha \geq 8 \left( \frac{\pi}{2} - \tan^{-1} \alpha \right) \]

This leads to:

\[ \tan^{-1} \alpha \geq \frac{8\pi}{10} \quad \Rightarrow \quad \alpha \geq \sqrt{3} \]

### Step 8: Find the minimum integral value
Since \( \sqrt{3} \approx 1.732 \), the minimum integral value of \( \alpha \) is:

\[ \alpha = 2 \]

### Conclusion
Thus, the minimum integral value of \( \alpha \) for which the quadratic equation has both positive roots is:

\[ \boxed{2} \]