If 2tan^2θ − 5secθ = 1 has exactly 7 solutions in the interval [0, nπ/2], for the least value of

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 \(n \in N\) then \(\sum\limits_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}}{2^{\mathrm{k}}}\) is equal to 

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

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

(3) \(1-\frac{15}{2^{13}}\)

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

Answer 1

Correct option is (4) \(\frac{1}{2^{13}}\left(2^{14}-15\right)\)

\(2 \tan ^{2} \theta-5 \sec \theta-1=0\)

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

\(\Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0\)

\(\Rightarrow \sec \theta=-\frac{1}{2}, 3\)

\(\Rightarrow \cos \theta=-2, \frac{1}{3}\)

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

For 7 solutions n = 13

So, \(\sum\limits_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S}\) (say)

\(\mathrm{S}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots .+\frac{13}{2^{13}}\)

\(\frac{1}{2} \mathrm{~S}=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}}\)

\(\Rightarrow \frac{S}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \)

\(\Rightarrow S=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}\)