Question

If \( 2 \tan^2 \theta - 5 \sec \theta = 1 \) has exactly 7 solutions in the interval \(\left[ 0, \frac{n\pi}{2} \right],\) for the least value of \( n \in \mathbb{N} \) then \(\sum_{k=1}^{n} \frac{k}{2^k}\) is equal to:

• A. \( \frac{1}{2^{15}} (2^{14} - 14) \)
• B. \( \frac{1}{2^{13}} (2^{14} - 15) \)
• C. \( \frac{1}{2^{14}} (2^{15} - 15) \)
• D. \( 1 - \frac{15}{2^{13}} \)

Answer 1

The Correct Option is B

To solve the given equation \( 2 \tan^2 \theta - 5 \sec \theta = 1 \) and find the least value of \( n \) such that this equation has exactly 7 solutions in the interval \( \left[ 0, \frac{n\pi}{2} \right] \), start by analyzing the trigonometric identity:

We know that \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan^2 \theta = \sec^2 \theta - 1\). Therefore, the equation can be expressed as:
\(2 (\sec^2 \theta - 1) - 5 \sec \theta = 1\)

Simplifying gives: 

\(2 \sec^2 \theta - 2 - 5 \sec \theta = 1\)

\(2 \sec^2 \theta - 5 \sec \theta - 3 = 0\)

Let \(x = \sec \theta\), making the equation:

\(2x^2 - 5x - 3 = 0\)

Solving this quadratic equation using the quadratic formula gives:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Plugging in the values where \( a = 2 \), \( b = -5 \), \( c = -3 \):

\(x = \frac{5 \pm \sqrt{25 + 24}}{4}\)

\(x = \frac{5 \pm \sqrt{49}}{4}\)

\(x = \frac{5 \pm 7}{4}\)

This results in two potential solutions for \( x \):

1. \(x = \frac{5 + 7}{4} = 3\)\)
2. \(x = \frac{5 - 7}{4} = -\frac{1}{2}\)\)

Since \( \sec \theta = x \), for real angles \( \theta \), only \( x = 3 \) is valid (as \( \sec \theta \) should be positive and refers to a value greater than or equal to 1 or less than or equal to -1).

For \(x = 3\)\), we have:

\(\sec \theta = 3 \Rightarrow \cos \theta = \frac{1}{3}\)\)

The general solution for \( \cos \theta = \frac{1}{3} \) within \( \theta \) cycle of \( 0 \) to \( 2\pi \) is:

\(\theta = 2k\pi \pm \cos^{-1}\left(\frac{1}{3}\right)\), where \( k \) is an integer.

The function \( \cos \theta = \frac{1}{3} \) will repeat for every \( 2\pi \) interval.

Now, find the smallest \( n \) such that there are exactly 7 solutions to \( \theta = \pm \cos^{-1}(\frac{1}{3}) \) in the interval \( \left[0, \frac{n\pi}{2} \right] \).

Each period from \( k = 0, 1, 2,... \) contains 2 solutions. We need at least 4 periods such that:

The interval length \( \frac{n\pi}{2} \geq 4\pi \) (for 4 full cycles, as every cycle causes 2 solutions).

Thus, \(n\frac{\pi}{2} = 4\pi\) implies:

\(\frac{n}{2} = 4 \Rightarrow n = 8\)

Finally, calculate:

\(\sum_{k=1}^{8} \frac{k}{2^k}\)

Using the formula for such series: \(S = \sum_{k=1}^{n} \frac{k}{2^k} = \frac{1}{2^{n+1}} (2^{n+1} - n - 1)\)

This implies:

\(\sum_{k=1}^{8} \frac{k}{2^k} = \frac{1}{2^{9}} (2^{9} - 8 - 1) = \frac{1}{2^{9}} (512 - 9)\)

So, the final answer matches the option:

\( \frac{1}{2^{13}} (2^{14} - 15) \)

Answer 2

Given: 
\(2\tan^2\theta - 5\sec\theta - 1 = 0.\)

Rewriting:
\(2\sec^2\theta - 5\sec\theta - 3 = 0.\)

Factoring:
\((2\sec\theta + 1)(\sec\theta - 3) = 0.\)

Thus:
\(\sec\theta = -\frac{1}{2}, \quad \sec\theta = 3.\)

Since \(\sec\theta = \frac{1}{\cos\theta}\), we find:  
\(\cos\theta = -2, \quad \cos\theta = \frac{1}{3}.\)

However, \(\cos\theta = -2\) is not possible, so:  
\(\cos\theta = \frac{1}{3}.\)

For the series sum:
Given:
\(S = \sum_{k=1}^{13} \frac{k}{2^k}.\)

The sum can be written as:  
\(S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \dots + \frac{13}{2^{13}}.\)

Multiplying the entire series by \(\frac{1}{2}\):
\(\frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}.\)

Subtracting:
\(S - \frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^{13}} - \frac{13}{2^{14}}.\)

Simplifying:
\(\frac{S}{2} = \left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{14}}.\)

Thus
\(S = 2\left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{13}}.\)

Simplifying further:
\(S = \frac{1}{2^{13}}(2^{14} - 15).\)

The correct option is (B) : \( \frac{1}{2^{13}} (2^{14} - 15) \)