Trigonometric Equations
Counting Solutions — Sum of Series
nta_pyq_2024_jan
Grade 11
Question:
If $2\tan^2\theta-5\sec\theta=1$ has exactly 7 solutions in the interval $\left[0,\dfrac{n\pi}{2}\right]$, for the least value of $n\in\mathbb{N}$, then $\displaystyle\sum_{k=1}^{n}\dfrac{k}{2^k}$ is equal to:
$\dfrac{1}{2^{10}}(2^{14}-14)$
$\dfrac{1}{2^{14}}(2^{15}-15)$
$1-\dfrac{15}{2^{13}}$
$\dfrac{1}{2^{13}}(2^{14}-15)$
Step-by-Step Solution
Key Concept: Solve: $2\sec^2\theta-5\sec\theta-3=0\Rightarrow(2\sec\theta+1)(\sec\theta-3)=0\Rightarrow\cos\theta=1/3$ (only valid). $\cos\theta=1/3$ has solutions at $\theta=\pm\cos^{-1}(1/3)+2k\pi$. Count 7 solutions in $[0,n\pi/2]$. Least $n=13$. Then compute $\sum_{k=1}^{13}k/2^k$.
$n=13$. $\sum_{k=1}^{13}k/2^k=\frac{2^{14}-15}{2^{13}}$.
Correct Answer: 4