Find all the values of the parameter'a' for which the function; `f(x)=8ax-a sin 6x-7x-sin 5x` increases & has no critical points for all `x in R`.

Answer

To solve the problem of finding all values of the parameter \( a \) for which the function

\[ f(x) = 8ax - a \sin(6x) - 7x - \sin(5x) \]

is increasing and has no critical points for all \( x \in \mathbb{R} \), we follow these steps:

### Step 1: Find the derivative of \( f(x) \)

To determine where the function is increasing, we first need to find the derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(8ax) - \frac{d}{dx}(a \sin(6x)) - \frac{d}{dx}(7x) - \frac{d}{dx}(\sin(5x)) \]

Calculating each term:

- The derivative of \( 8ax \) is \( 8a \).
- The derivative of \( a \sin(6x) \) is \( 6a \cos(6x) \).
- The derivative of \( 7x \) is \( 7 \).
- The derivative of \( \sin(5x) \) is \( 5 \cos(5x) \).

Thus, we have:

\[ f'(x) = 8a - 6a \cos(6x) - 7 - 5 \cos(5x) \]

### Step 2: Set the derivative greater than zero

For the function \( f(x) \) to be increasing, we require:

\[ f'(x) > 0 \]

This leads to:

\[ 8a - 6a \cos(6x) - 7 - 5 \cos(5x) > 0 \]

### Step 3: Rearranging the inequality

Rearranging the inequality gives:

\[ 8a - 7 - 6a \cos(6x) - 5 \cos(5x) > 0 \]

### Step 4: Analyze the maximum values of cosine functions

The terms \( \cos(6x) \) and \( \cos(5x) \) can take values between -1 and 1. To ensure that the inequality holds for all \( x \), we need to consider the worst-case scenario where \( \cos(6x) \) and \( \cos(5x) \) take their maximum values (which is 1):

\[ 8a - 7 - 6a - 5 > 0 \]

This simplifies to:

\[ 2a - 12 > 0 \]

### Step 5: Solve for \( a \)

Solving the inequality:

\[ 2a > 12 \implies a > 6 \]

### Conclusion

Thus, the function \( f(x) \) is increasing and has no critical points for all \( x \in \mathbb{R} \) when:

\[ a > 6 \]