f(x)=x^3 + 3x^2 + 4x + a(sinx) + b(cosx) for all x belongs to R is a injective function den d value of a^2 + b^2 is:

To determine the values of \( a \) and \( b \) such that the function \( f(x) = x^3 + 3x^2 + 4x + a \sin x + b \cos x \) is injective for all \( x \in \mathbb{R} \), we need to analyze the conditions under which a function is injective. A function is injective if it is either strictly increasing or strictly decreasing across its entire domain.

Understanding the Derivative

The first step in assessing whether \( f(x) \) is injective is to compute its derivative:

\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 + 4x + a \sin x + b \cos x) = 3x^2 + 6x + 4 + a \cos x - b \sin x \]

Analyzing the Derivative

For \( f(x) \) to be injective, \( f'(x) \) must not change sign. This means we need to ensure that \( f'(x) \) is either always positive or always negative. The polynomial part \( 3x^2 + 6x + 4 \) is a quadratic function that opens upwards (since the coefficient of \( x^2 \) is positive). We can analyze its discriminant to see if it has any real roots:

\[ D = b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12 \]

Since the discriminant is negative, the quadratic \( 3x^2 + 6x + 4 \) has no real roots and is always positive. Therefore, \( 3x^2 + 6x + 4 > 0 \) for all \( x \in \mathbb{R} \).

Incorporating the Trigonometric Terms

Now we need to consider the terms \( a \cos x - b \sin x \). The maximum and minimum values of \( a \cos x - b \sin x \) can be found using the amplitude of the expression. The maximum value occurs when \( a \) and \( b \) are aligned in the same direction:

\[ \sqrt{a^2 + b^2} \]

Thus, the range of \( a \cos x - b \sin x \) is from \( -\sqrt{a^2 + b^2} \) to \( \sqrt{a^2 + b^2} \).

Ensuring Positivity of the Derivative

For \( f'(x) \) to remain positive, we need:

\[ 3x^2 + 6x + 4 + a \cos x - b \sin x > 0 \]

Since \( 3x^2 + 6x + 4 \) is always positive, we need to ensure that the minimum value of \( a \cos x - b \sin x \) does not bring \( f'(x) \) to zero or below. Therefore, we require:

\[ -\sqrt{a^2 + b^2} > - (3x^2 + 6x + 4) \]

To satisfy this for all \( x \), we can set the minimum of \( 3x^2 + 6x + 4 \) (which is \( 4 - 3 \) at \( x = -1 \)) to be greater than \( -\sqrt{a^2 + b^2} \). Thus, we need:

\[ \sqrt{a^2 + b^2} < 4 \]

Finding the Value of \( a^2 + b^2 \)

Squaring both sides gives us:

\[ a^2 + b^2 < 16 \]

To find the maximum value of \( a^2 + b^2 \) that keeps the function injective, we can conclude that the largest possible value is \( 16 \). Therefore, the answer to the problem is:

\[ \boxed{16} \]