If k be positive integer , then the number of values of k satisfying `int_0^(pi/2)[k^2(cos3x+1/2cosx)+sinx-2kcosx]dxlt=1` is

Options

A. 10
B. 11
C. 12
D. none of these

Answer

To solve the problem, we need to evaluate the integral and find the number of positive integer values of \( k \) that satisfy the inequality:

\[ \int_0^{\frac{\pi}{2}} \left[ k^2 \left( \cos(3x) + \frac{1}{2} \cos(x) \right) + \sin(x) - 2k \cos(x) \right] dx < 1 \]

### Step 1: Break down the integral

We can rewrite the integral as:

\[ \int_0^{\frac{\pi}{2}} \left[ k^2 \cos(3x) + \frac{k^2}{2} \cos(x) + \sin(x) - 2k \cos(x) \right] dx \]

### Step 2: Evaluate each term separately

1. **Integral of \( k^2 \cos(3x) \)**:
\[ \int_0^{\frac{\pi}{2}} k^2 \cos(3x) \, dx = k^2 \left[ \frac{\sin(3x)}{3} \right]_0^{\frac{\pi}{2}} = k^2 \left( \frac{\sin\left(\frac{3\pi}{2}\right) - \sin(0)}{3} \right) = k^2 \left( \frac{-1 - 0}{3} \right) = -\frac{k^2}{3} \]

2. **Integral of \( \frac{k^2}{2} \cos(x) \)**:
\[ \int_0^{\frac{\pi}{2}} \frac{k^2}{2} \cos(x) \, dx = \frac{k^2}{2} \left[ \sin(x) \right]_0^{\frac{\pi}{2}} = \frac{k^2}{2} (1 - 0) = \frac{k^2}{2} \]

3. **Integral of \( \sin(x) \)**:
\[ \int_0^{\frac{\pi}{2}} \sin(x) \, dx = \left[ -\cos(x) \right]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

4. **Integral of \( -2k \cos(x) \)**:
\[ \int_0^{\frac{\pi}{2}} -2k \cos(x) \, dx = -2k \left[ \sin(x) \right]_0^{\frac{\pi}{2}} = -2k (1 - 0) = -2k \]

### Step 3: Combine the results

Now combine all the evaluated integrals:

\[ -\frac{k^2}{3} + \frac{k^2}{2} + 1 - 2k \]

### Step 4: Simplify the expression

To combine the terms involving \( k^2 \):

\[ -\frac{k^2}{3} + \frac{k^2}{2} = \frac{-2k^2 + 3k^2}{6} = \frac{k^2}{6} \]

Thus, the entire expression becomes:

\[ \frac{k^2}{6} + 1 - 2k \]

### Step 5: Set up the inequality

Now we need to satisfy the inequality:

\[ \frac{k^2}{6} + 1 - 2k < 1 \]

Subtracting 1 from both sides gives:

\[ \frac{k^2}{6} - 2k < 0 \]

### Step 6: Factor the inequality

Multiplying through by 6 (since 6 is positive, the inequality direction remains unchanged):

\[ k^2 - 12k < 0 \]

Factoring out \( k \):

\[ k(k - 12) < 0 \]

### Step 7: Determine the intervals

The critical points are \( k = 0 \) and \( k = 12 \). The inequality \( k(k - 12) < 0 \) is satisfied in the interval:

\[ 0 < k < 12 \]

### Step 8: Find positive integer solutions

The positive integer values of \( k \) that satisfy this inequality are:

\[ k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \]

### Conclusion

Thus, there are **11 positive integer values** of \( k \) that satisfy the given inequality.