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 solution for `x in (0,1)`

Answer

To solve the question, we will analyze each of the given statements one by one, using calculus concepts such as the first and second derivatives to determine the number of real solutions for the equations provided.

### Step-by-Step Solution:

1. **Analyzing the first statement:**
\[ f(x) = x^4 + 2x^2 - 6x + 2 = 0 \]
- Calculate the first derivative:
\[ f'(x) = 4x^3 + 4x - 6 \]
- Calculate the second derivative:
\[ f''(x) = 12x^2 + 4 \]
- Since \(f''(x) > 0\) for all \(x\) (as \(12x^2 + 4\) is always positive), \(f'(x)\) is strictly increasing. This implies that \(f'(x)\) can have at most one real root.
- Therefore, \(f(x)\) can have at most two real roots.
- **Conclusion:** The first statement is **incorrect**.

2. **Analyzing the second statement:**
\[ g(x) = x^3 + 5x + 1 = 0 \]
- Calculate the first derivative:
\[ g'(x) = 3x^2 + 5 \]
- Since \(g'(x) > 0\) for all \(x\) (as \(3x^2 + 5\) is always positive), \(g(x)\) is strictly increasing.
- A strictly increasing function can cross the x-axis at most once, meaning there is exactly one real root.
- **Conclusion:** The second statement is **incorrect**.

3. **Analyzing the third statement:**
\[ h(x) = x^n + ax + b = 0 \quad (n \text{ is even}) \]
- If \(a = 0\), then:
\[ h(x) = x^n + b = 0 \Rightarrow x^n = -b \]
- This gives two solutions \(x = \pm (-b)^{1/n}\) if \(b < 0\).
- If \(h(x)\) has more than two solutions, then by Rolle's Theorem, \(h'(x)\) must have at least \(n-1\) roots. However, since \(h'(x) = nx^{n-1} + a\) can have at most one root (as \(n\) is even), this leads to a contradiction.
- **Conclusion:** The third statement is **correct**.

4. **Analyzing the fourth statement:**
\[ k(x) = x^3 - 3x + c = 0 \]
- Set \(c = 0\):
\[ k(x) = x^3 - 3x = 0 \Rightarrow x(x^2 - 3) = 0 \Rightarrow x = 0, \pm\sqrt{3} \]
- The roots \(0, -\sqrt{3}, \sqrt{3}\) do not lie in the interval \((0, 1)\).
- Therefore, there are no solutions in the interval \(0 < x < 1\).
- **Conclusion:** The fourth statement is **incorrect**.

### Final Conclusion:
- The only correct statement is the third one.