The least positive integral value of 'x' satisfying `(e^x-2)(sin(x+pi/4))(x-log_e2)(sinx-cosx)<0`

Answer

To find the least positive integral value of \( x \) satisfying the inequality

\[ (e^x - 2)(\sin(x + \frac{\pi}{4})) (x - \log_e 2)(\sin x - \cos x) < 0, \]

we will analyze each factor in the product step by step.

### Step 1: Analyze \( e^x - 2 \)

Set \( e^x - 2 = 0 \):

\[ e^x = 2 \implies x = \log_e 2 \approx 0.693. \]

- For \( x < \log_e 2 \) (approximately \( 0.693 \)), \( e^x - 2 < 0 \).
- For \( x > \log_e 2 \), \( e^x - 2 > 0 \).

### Step 2: Analyze \( x - \log_e 2 \)

Set \( x - \log_e 2 = 0 \):

\[ x = \log_e 2 \approx 0.693. \]

- For \( x < \log_e 2 \), \( x - \log_e 2 < 0 \).
- For \( x > \log_e 2 \), \( x - \log_e 2 > 0 \).

### Step 3: Combine \( e^x - 2 \) and \( x - \log_e 2 \)

Both \( e^x - 2 \) and \( x - \log_e 2 \) are negative for \( x < \log_e 2 \) and positive for \( x > \log_e 2 \). Therefore, their product is:

\[ (e^x - 2)(x - \log_e 2) \geq 0 \text{ for all } x \neq \log_e 2. \]

### Step 4: Analyze \( \sin(x + \frac{\pi}{4}) \)

The function \( \sin(x + \frac{\pi}{4}) \) oscillates between -1 and 1. We need to find where it is negative:

- \( \sin(x + \frac{\pi}{4}) < 0 \) occurs in intervals of the form \( (n\pi - \frac{\pi}{4}, n\pi + \frac{3\pi}{4}) \) for integers \( n \).

### Step 5: Analyze \( \sin x - \cos x \)

Set \( \sin x - \cos x = 0 \):

\[ \tan x = 1 \implies x = \frac{\pi}{4} + n\pi \text{ for integers } n. \]

- For \( x \) in intervals \( \left( \frac{\pi}{4} + n\pi, \frac{5\pi}{4} + n\pi \right) \), \( \sin x - \cos x < 0 \).

### Step 6: Combine all factors

We need to find intervals where the product

\[ (e^x - 2)(\sin(x + \frac{\pi}{4}))(x - \log_e 2)(\sin x - \cos x) < 0. \]

From our analysis:

1. \( e^x - 2 \) and \( x - \log_e 2 \) are both positive for \( x > \log_e 2 \).
2. \( \sin(x + \frac{\pi}{4}) \) is negative in certain intervals.
3. \( \sin x - \cos x \) is negative in certain intervals.

### Step 7: Find the least positive integral value of \( x \)

We need to find the least positive integer \( x \) such that the product is negative.

- Check \( x = 1 \):
- \( e^1 - 2 > 0 \)
- \( 1 - \log_e 2 > 0 \)
- \( \sin(1 + \frac{\pi}{4}) > 0 \)
- \( \sin 1 - \cos 1 < 0 \) (approximately)

The product is positive.

- Check \( x = 2 \):
- \( e^2 - 2 > 0 \)
- \( 2 - \log_e 2 > 0 \)
- \( \sin(2 + \frac{\pi}{4}) < 0 \)
- \( \sin 2 - \cos 2 < 0 \)

The product is negative.

- Check \( x = 3 \):
- \( e^3 - 2 > 0 \)
- \( 3 - \log_e 2 > 0 \)
- \( \sin(3 + \frac{\pi}{4}) < 0 \)
- \( \sin 3 - \cos 3 < 0 \)

The product is negative.

Thus, the least positive integral value of \( x \) satisfying the inequality is:

\[ \boxed{2}. \]