Statement-1: The function `f:R to R` defined by `f(x)=x^(3)+4x-5` is a bijection.
Statement-2: Every odd degree has at least one real root.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the statements given regarding the function \( f(x) = x^3 + 4x - 5 \).

### Step 1: Determine if \( f(x) \) is a bijection
A function is a bijection if it is both injective (one-to-one) and surjective (onto).

#### Injectivity:
To check if \( f(x) \) is injective, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ x_1^3 + 4x_1 - 5 = x_2^3 + 4x_2 - 5 \]
This simplifies to:
\[ x_1^3 - x_2^3 + 4(x_1 - x_2) = 0 \]
Factoring gives:
\[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2 + 4) = 0 \]
This implies either \( x_1 - x_2 = 0 \) (which means \( x_1 = x_2 \)) or \( x_1^2 + x_1x_2 + x_2^2 + 4 = 0 \).

The term \( x_1^2 + x_1x_2 + x_2^2 + 4 \) is always positive since \( x_1^2 + x_1x_2 + x_2^2 \) is non-negative (as it can be rewritten using the squares of real numbers) and adding 4 ensures it is strictly greater than 0. Thus, \( f(x) \) is injective.

#### Surjectivity:
Next, we check if \( f(x) \) is surjective. Since \( f(x) \) is a polynomial of odd degree (3), it will cover all real numbers as \( x \) approaches \( \pm \infty \). Therefore, \( f(x) \) is surjective.

Since \( f(x) \) is both injective and surjective, it is a bijection.

### Step 2: Verify Statement 2
Statement 2 states that every odd degree polynomial has at least one real root. Since \( f(x) \) is a cubic polynomial (odd degree), it must have at least one real root. This statement is true.

### Conclusion
Both statements are true. Therefore, the answer is that both statements are valid.

### Final Answer
Both Statement-1 and Statement-2 are true.

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