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Mathematics Question on Derivatives of Functions in Parametric Forms

If 2tan⁡2θ−5sec⁡θ=12 \tan^2 \theta - 5 \sec \theta = 1 has exactly 7 solutions in the interval [0,nπ2],\left[ 0, \frac{n\pi}{2} \right], for the least value of n∈Nn \in \mathbb{N} then ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k} is equal to:

A

1215(214−14)\frac{1}{2^{15}} (2^{14} - 14)

B

1213(214−15)\frac{1}{2^{13}} (2^{14} - 15)

C

1214(215−15)\frac{1}{2^{14}} (2^{15} - 15)

D

1−152131 - \frac{15}{2^{13}}

Answer

1213(214−15)\frac{1}{2^{13}} (2^{14} - 15)

Explanation

Solution

Given:
2tan⁡2θ−5sec⁡θ−1=0.2\tan^2\theta - 5\sec\theta - 1 = 0.

Rewriting:
2sec⁡2θ−5sec⁡θ−3=0.2\sec^2\theta - 5\sec\theta - 3 = 0.

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

Thus:
sec⁡θ=−12,sec⁡θ=3.\sec\theta = -\frac{1}{2}, \quad \sec\theta = 3.

Since sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, we find:
cos⁡θ=−2,cos⁡θ=13.\cos\theta = -2, \quad \cos\theta = \frac{1}{3}.

However, cos⁡θ=−2\cos\theta = -2 is not possible, so:
cos⁡θ=13.\cos\theta = \frac{1}{3}.

For the series sum:
Given:
S=∑k=113k2k.S = \sum_{k=1}^{13} \frac{k}{2^k}.

The sum can be written as:
S=12+222+323+⋯+13213.S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \dots + \frac{13}{2^{13}}.

Multiplying the entire series by 12\frac{1}{2}:
S2=122+223+⋯+12213+13214.\frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}.

Subtracting:
S−S2=12+122+123+⋯+1213−13214.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:
S2=(1−1213)−13214.\frac{S}{2} = \left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{14}}.

Thus
S=2(1−1213)−13213.S = 2\left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{13}}.

Simplifying further:
S=1213(214−15).S = \frac{1}{2^{13}}(2^{14} - 15).

The correct option is (B) : 1213(214−15)\frac{1}{2^{13}} (2^{14} - 15)