If 2tan²(θ) - 5sec(θ) = 1 has exactly 7 solutions in the interval… | JEE Main 2024 Mathematics PYQ

MCQMediumJEE Main 2024 · 27 January, Shift 2Trigonometric Equations

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

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