JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(n \in N\) then \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}}{2^{\mathrm{k}}}\) is equal to :
- A \(\frac{1}{2^{15}}\left(2^{14}-14\right)\)
- B \(\frac{1}{2^{14}}\left(2^{15}-15\right)\)
- C \(1-\frac{15}{2^{13}}\)
- D \(\frac{1}{2^{13}}\left(2^{14}-15\right)\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2^{13}}\left(2^{14}-15\right)\)
Step-by-step Solution
Detailed explanation
\(2 \tan ^2 \theta-5 \sec \theta-1=0 \) \( \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \) \( \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \) \( \Rightarrow \sec \theta=-\frac{1}{2}, \) \( \Rightarrow \cos \theta=-2, \frac{1}{3} \) \( \Rightarrow \cos \theta=\frac{1}{3}\) For…
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