If `2tan^2theta-5sectheta=1` has exactly 7 solutions in `[0,(npi)/2]` for least value of `ninN`, then `sum_(k=1)^nk/(2n)` is equal to..................

Options

A. `9/2^9`
B. `9^1/2^13`
C. `7/2^7`
D. `11/2^12`

Answer

To solve the equation \( 2\tan^2\theta - 5\sec\theta = 1 \) and find the least value of \( n \) such that the equation has exactly 7 solutions in the interval \( [0, \frac{n\pi}{2}] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2\tan^2\theta - 5\sec\theta = 1 \]
Using the identity \( \tan^2\theta = \sec^2\theta - 1 \), we can rewrite \( \tan^2\theta \):
\[ 2(\sec^2\theta - 1) - 5\sec\theta = 1 \]
This simplifies to:
\[ 2\sec^2\theta - 2 - 5\sec\theta = 1 \]
Rearranging gives:
\[ 2\sec^2\theta - 5\sec\theta - 3 = 0 \]

### Step 2: Solve the quadratic equation
Now we will solve the quadratic equation \( 2\sec^2\theta - 5\sec\theta - 3 = 0 \) using the quadratic formula:
\[ \sec\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -5, c = -3 \):
\[ \sec\theta = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \sec\theta = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4} \]
This gives us two solutions:
\[ \sec\theta = \frac{12}{4} = 3 \quad \text{and} \quad \sec\theta = \frac{-2}{4} = -\frac{1}{2} \]

### Step 3: Determine valid solutions
Since \( \sec\theta \) cannot be negative in the interval \( [0, \frac{n\pi}{2}] \), we discard \( \sec\theta = -\frac{1}{2} \). Thus, we only consider:
\[ \sec\theta = 3 \implies \cos\theta = \frac{1}{3} \]

### Step 4: Find the number of solutions
The equation \( \cos\theta = \frac{1}{3} \) has solutions in the interval \( [0, \frac{n\pi}{2}] \). The cosine function is positive in the first quadrant and decreases to zero at \( \frac{\pi}{2} \). Therefore, there are two solutions in each interval of \( [0, \pi] \) and \( [\pi, 2\pi] \), and so on.

To find the number of solutions:
- From \( 0 \) to \( 2\pi \): 2 solutions
- From \( 2\pi \) to \( 4\pi \): 2 solutions
- From \( 4\pi \) to \( 6\pi \): 2 solutions
- From \( 6\pi \) to \( 13\pi/2 \): 1 solution

Thus, we have:
\[ 2 + 2 + 2 + 1 = 7 \text{ solutions} \]
This means \( n = 13 \).

### Step 5: Calculate the sum
Now we need to compute:
\[ \sum_{k=1}^{n} \frac{k}{2n} = \frac{1}{2n} \sum_{k=1}^{n} k \]
Using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
Substituting \( n = 13 \):
\[ \sum_{k=1}^{13} k = \frac{13 \cdot 14}{2} = 91 \]
Thus:
\[ \sum_{k=1}^{13} \frac{k}{2 \cdot 13} = \frac{91}{26} = \frac{91}{26} = 3.5 \]

### Final Answer
The value of \( \sum_{k=1}^{n} \frac{k}{2n} \) is:
\[ \frac{91}{26} \]