If the roots of the equation, `x^3 + px^2+qx-1 = 0` form an increasing` G.P.` where `p` and `q`are real,then (a) `p +q = 0` (b) `p in (-3, oo)` (c) one of the roots is unity (d) one root is smaller than 1 and one root is greater than 1

Options

A. `p +q = 0`
B. `pin (-3, oo)`
C. one of the roots is untiy
D. one roots is smaller than 1 and one roots is greater than 1

Explanation

To solve the problem, we need to analyze the polynomial equation \(x^3 + px^2 + qx - 1 = 0\) given that its roots form an increasing geometric progression (G.P.). Let's denote the roots as \(a\), \(ar\), and \(ar^2\) where \(a\) is the first term and \(r > 1\) is the common ratio.

### Step 1: Sum of the Roots
According to Vieta's formulas, the sum of the roots of the polynomial is given by:
\[ a + ar + ar^2 = -p \]
Factoring out \(a\), we have:
\[ a(1 + r + r^2) = -p \tag{1} \]

### Step 2: Sum of the Product of the Roots Taken Two at a Time
The sum of the products of the roots taken two at a time is:
\[ a \cdot ar + a \cdot ar^2 + ar \cdot ar^2 = q \]
This simplifies to:
\[ a^2r + a^2r^2 + a^2r^3 = q \]
Factoring out \(a^2\), we get:
\[ a^2r(1 + r + r^2) = q \tag{2} \]

### Step 3: Product of the Roots
The product of the roots is given by:
\[ a \cdot ar \cdot ar^2 = a^3r^3 = -(-1) = 1 \]
Thus, we have:
\[ a^3r^3 = 1 \implies ar = 1 \tag{3} \]

### Step 4: Relating \(p\) and \(q\)
From equation (3), we can substitute \(ar = 1\) into equations (1) and (2):
- From (1):
\[ a(1 + r + r^2) = -p \]
Substituting \(a = \frac{1}{r}\) (since \(ar = 1\)):
\[ \frac{1}{r}(1 + r + r^2) = -p \implies 1 + r + r^2 = -pr \]
Thus, we have:
\[ p = -\frac{1 + r + r^2}{r} \tag{4} \]

- From (2):
Using \(ar = 1\):
\[ a^2r(1 + r + r^2) = q \implies \left(\frac{1}{r^2}\right)r(1 + r + r^2) = q \implies \frac{1 + r + r^2}{r} = q \tag{5} \]

### Step 5: Finding the Relationship Between \(p\) and \(q\)
From equations (4) and (5), we can see:
\[ q = -p \]
Thus, we have:
\[ p + q = 0 \tag{6} \]

### Step 6: Analyzing the Roots
Since \(ar = 1\), one of the roots is \(1\). The roots are \(a\), \(1\), and \(ar^2\). Since \(r > 1\), we have:
- \(a < 1\) (because \(ar = 1\))
- \(ar^2 > 1\) (because \(r^2 > 1\))

This means:
- One root is \(1\) (unity).
- One root is less than \(1\).
- One root is greater than \(1\).

### Step 7: Finding the Range of \(p\)
From the expression for \(p\) in equation (4):
\[ p = -\frac{1 + r + r^2}{r} \]
Since \(r > 1\), the numerator \(1 + r + r^2\) is positive. Therefore, \(p\) will be negative. To analyze the range, we note that as \(r\) approaches \(1\), \(p\) approaches \(-3\). Hence, \(p\) must be greater than \(-3\):
\[ p > -3 \]

### Conclusion
From the analysis, we conclude:
- \(p + q = 0\) (Option a)
- \(p > -3\) (Option b)
- One root is \(1\) (Option c)
- One root is less than \(1\) and one root is greater than \(1\) (Option d)

Thus, all options (a), (b), (c), and (d) are correct.