Trigonometric Equations Question 3
Question 3 - 2024 (27 Jan Shift 2)
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 :
(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) $\frac{1}{2^{13}}\left(2^{14}-15\right)$
Show Answer
Answer (4)
Solution
$$ \begin{aligned} & 2 \tan ^{2} \theta-5 \sec \theta-1=0 \ & \Rightarrow 2 \sec ^{2} \theta-5 \sec \theta-3=0 \ & \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \ & \Rightarrow \sec \theta=-\frac{1}{2}, 3 \ & \Rightarrow \cos \theta=-2, \frac{1}{3} \ & \Rightarrow \cos \theta=\frac{1}{3} \end{aligned} $$
For 7 solutions $\mathrm{n}=13$
So, $\sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S}$ (say)
$\mathrm{S}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots+\frac{13}{2^{13}}$
$\frac{1}{2} S=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}}$
$\Rightarrow \frac{S}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow S=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}$
