(i) Find the values of 'a' for which the expression
`x^(2)-(3a-1)x+2a^(2)+2a-11` is always positive
(ii) If `x^(2)+4ax+2 gt0` for all values of x, then a lies in the interval.
(a) `(-2,4)`
(b) (1,2)
(c) `(-sqrt(2),sqrt(2))`
(d) `(-(1)/(sqrt(2)),(1)/(sqrt(2)))`
(e) (-4,2).

Answer

To solve the given problem step by step, we will break it down into two parts as per the question.

### Part (i): Finding values of 'a' for which the expression is always positive

We need to determine the values of \( a \) such that the expression

\[ x^2 - (3a - 1)x + (2a^2 + 2a - 11) \]

is always positive. For a quadratic expression \( ax^2 + bx + c \) to be always positive, its discriminant \( D \) must be less than zero.

1. **Identify the coefficients**:
- Here, \( a = 1 \) (coefficient of \( x^2 \)),
- \( b = -(3a - 1) = -3a + 1 \),
- \( c = 2a^2 + 2a - 11 \).

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-3a + 1)^2 - 4(1)(2a^2 + 2a - 11) \]

3. **Expand the discriminant**:
\[ D = (9a^2 - 6a + 1) - (8a^2 + 8a - 44) \]
Simplifying this:
\[ D = 9a^2 - 6a + 1 - 8a^2 - 8a + 44 \]
\[ D = a^2 - 14a + 45 \]

4. **Set the discriminant less than zero**:
\[ a^2 - 14a + 45 < 0 \]

5. **Factor the quadratic**:
We can factor this as:
\[ (a - 5)(a - 9) < 0 \]

6. **Find the intervals**:
The roots of the equation are \( a = 5 \) and \( a = 9 \). The quadratic will be negative between the roots:
\[ 5 < a < 9 \]

### Conclusion for Part (i):
The values of \( a \) for which the expression is always positive are:
\[ a \in (5, 9) \]

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### Part (ii): Finding the interval for \( a \) such that \( x^2 + 4ax + 2 > 0 \) for all \( x \)

1. **Identify the coefficients**:
- Here, \( a = 1 \),
- \( b = 4a \),
- \( c = 2 \).

2. **Calculate the discriminant**:
\[ D = (4a)^2 - 4(1)(2) \]
\[ D = 16a^2 - 8 \]

3. **Set the discriminant less than zero**:
\[ 16a^2 - 8 < 0 \]

4. **Simplify the inequality**:
\[ 16a^2 < 8 \]
\[ 2a^2 < 1 \]
\[ a^2 < \frac{1}{2} \]

5. **Find the interval for \( a \)**:
Taking square roots:
\[ -\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}} \]

### Conclusion for Part (ii):
The values of \( a \) for which \( x^2 + 4ax + 2 > 0 \) for all \( x \) are:
\[ a \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]

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