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 quadratic equation can be expressed in the standard form \( Ax^2 + Bx + C = 0 \), where:
- \( A = \cot^{-1} \alpha \)
- \( B = -(\tan^{-1} \alpha)^{3/2} \)
- \( C = 2(\cot^{-1} \alpha)^2 \)

### Step 2: Use the condition for positive roots
For a quadratic equation to have both roots positive, the following conditions must be satisfied:
1. \( A > 0 \)
2. \( C > 0 \)
3. The discriminant \( D = B^2 - 4AC \) must be non-negative, i.e., \( D \geq 0 \).

### Step 3: Check the conditions
1. **Condition \( A > 0 \)**:
- Since \( \cot^{-1} \alpha \) is positive for \( \alpha > 0 \), this condition is satisfied for \( \alpha > 0 \).

2. **Condition \( C > 0 \)**:
- \( C = 2(\cot^{-1} \alpha)^2 > 0 \) is also satisfied for \( \alpha > 0 \).

3. **Condition for the discriminant**:
- We need to ensure that \( D = B^2 - 4AC \geq 0 \).
- Calculate \( D \):
\[ D = \left(-(\tan^{-1} \alpha)^{3/2}\right)^2 - 4(\cot^{-1} \alpha)(2(\cot^{-1} \alpha)^2) \]
\[ = (\tan^{-1} \alpha)^{3} - 8(\cot^{-1} \alpha)^3 \]
- We need \( (\tan^{-1} \alpha)^{3} \geq 8(\cot^{-1} \alpha)^3 \).

### Step 4: Use the identity
Using the identity \( \tan^{-1} \alpha + \cot^{-1} \alpha = \frac{\pi}{2} \):
- Let \( \tan^{-1} \alpha = \theta \), then \( \cot^{-1} \alpha = \frac{\pi}{2} - \theta \).
- Substitute into the inequality:
\[ \theta^3 \geq 8\left(\frac{\pi}{2} - \theta\right)^3 \]

### Step 5: Solve the inequality
To find the minimum integral value of \( \alpha \):
1. We can find \( \alpha \) such that \( \tan^{-1} \alpha \geq \frac{\pi}{3} \) (since \( \tan(\frac{\pi}{3}) = \sqrt{3} \)).
2. Therefore, \( \alpha \geq \tan(\frac{\pi}{3}) = \sqrt{3} \).

### Step 6: Find the minimum integral value
The minimum integral value of \( \alpha \) greater than \( \sqrt{3} \) is \( 2 \).

### Final Answer
Thus, the minimum integral value of \( \alpha \) for which the quadratic equation has both positive roots is \( \boxed{2} \).

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