If f :R-R , where f(x) = ax +cosx and f(x) is bijective , thena belongs to Ra belongs to R+a belongs to R-a belongs to R - (-1,1)

To determine the values of \( a \) for which the function \( f(x) = ax + \cos(x) \) is bijective, we need to analyze both the injective (one-to-one) and surjective (onto) properties of the function. A function is bijective if it is both injective and surjective.

Understanding Injectivity

A function is injective if different inputs produce different outputs. For \( f(x) \) to be injective, its derivative must not change sign (i.e., it should either be always positive or always negative). Let's find the derivative of \( f(x) \):

The derivative is given by:

f'(x) = a - sin(x)

To ensure that \( f'(x) \) does not change sign, we analyze the expression \( a - \sin(x) \). The sine function oscillates between -1 and 1, so we can set up inequalities to find the conditions on \( a \):

• For \( f'(x) \) to be always positive: a - 1 > 0 or a > 1
• For \( f'(x) \) to be always negative: a + 1 < 0 or a < -1

Thus, \( f(x) \) is injective if \( a > 1 \) or \( a < -1 \).

Exploring Surjectivity

A function is surjective if it covers the entire range of real numbers. Since \( \cos(x) \) oscillates between -1 and 1, we need to consider how the linear term \( ax \) affects the overall behavior of \( f(x) \). As \( x \) approaches positive or negative infinity, the term \( ax \) will dominate the behavior of \( f(x) \). Therefore:

• If \( a > 0 \), as \( x \to \infty \), \( f(x) \to \infty \) and as \( x \to -\infty \), \( f(x) \to -\infty \).
• If \( a < 0 \), as \( x \to \infty \), \( f(x) \to -\infty \) and as \( x \to -\infty \), \( f(x) \to \infty \).

In both cases, \( f(x) \) can cover all real numbers, ensuring surjectivity.

Combining Conditions for Bijectivity

To summarize, for \( f(x) = ax + \cos(x) \) to be bijective, we need:

• Injective: \( a > 1 \) or \( a < -1 \)
• Surjective: \( a \) can be any real number, but the injective condition must hold.

Thus, the values of \( a \) that make \( f(x) \) bijective are:

a \in (-\infty, -1) \cup (1, \infty)

In conclusion, the function \( f(x) = ax + \cos(x) \) is bijective for values of \( a \) that lie outside the interval \([-1, 1]\), specifically when \( a < -1 \) or \( a > 1\). This ensures that the function is both injective and surjective, fulfilling the requirements for bijectivity.