Which of the following are correct

Options

A. `x ^(4) + 2x ^(2) - 6x+2=0` has exactly four real solution
B. `x ^(3) + 5x +1=0` has exactly three real solutions
C. `x ^(n) +ax +b=0`where n is an even natural number has atmost two real solution `a,b,`in R.
D. `x ^(3)-3x+c=0, x gt 0` has two real solutin for `x in (0,1)`

Answer

To solve the question, we will analyze each option one by one and determine their correctness based on the properties of polynomials and their derivatives.

### Step-by-Step Solution:

**Option 1: \(x^4 + 2x^2 - 6x + 2\) has exactly 4 solutions.**

1. **Find the first derivative:**
\[ f'(x) = 4x^3 + 4x - 6 \]

2. **Find the second derivative:**
\[ f''(x) = 12x^2 + 4 \]
Since \(f''(x)\) is always positive (as \(12x^2 + 4 > 0\) for all \(x\)), the function is always concave up.

3. **Conclusion:**
A polynomial that is always concave up can have at most 2 real roots. Therefore, this option is **incorrect**.

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**Option 2: \(x^3 + 5x + 1 = 0\) has exactly 3 solutions.**

1. **Find the first derivative:**
\[ f'(x) = 3x^2 + 5 \]
Since \(f'(x)\) is always positive (as \(3x^2 + 5 > 0\) for all \(x\)), the function is strictly increasing.

2. **Conclusion:**
A strictly increasing function can cross the x-axis at most once. Therefore, this option is **incorrect**.

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**Option 3: \(x^n + ax + b\) (where \(n\) is an even natural number) has at most two real solutions.**

1. **Find the first derivative:**
\[ f'(x) = nx^{n-1} + a \]

2. **Find the second derivative:**
\[ f''(x) = n(n-1)x^{n-2} \]
Since \(n\) is even, \(f''(x)\) is always non-negative, indicating that the function is either always increasing or has a minimum point.

3. **Conclusion:**
This means the function can have at most 2 real roots. Therefore, this option is **correct**.

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**Option 4: \(x^3 - 3x + c = 0\) has two real solutions for \(x \in (0, 1)\).**

1. **Evaluate the function at the endpoints:**
\[ f(0) = c \]
\[ f(1) = 1 - 3 + c = c - 2 \]

2. **Check the signs:**
For the function to have two solutions in the interval (0, 1), \(f(0)\) and \(f(1)\) must have opposite signs:
\[ f(0) \cdot f(1) < 0 \implies c(c - 2) < 0 \]
This inequality holds for \(0 < c < 2\).

3. **Conclusion:**
This means that depending on the value of \(c\), the function can indeed have two real solutions in the interval (0, 1). However, this is not true for all \(c\). Therefore, this option is **incorrect**.

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### Final Conclusion:
- **Correct Option:** Option 3 is correct.
- **Incorrect Options:** Options 1, 2, and 4 are incorrect.