Jee Main 2024 27 01 2024 Shift 2 - Question2
Question 2
If $2 \tan ^{2} \theta-5 \sec \theta=1$ has exactly 7 solutions in $\left[0, \frac{n \pi}{2}\right]$ for least value of $n \in N$, then $\sum _{K=1}^{n} \frac{K}{2^{n}}$ is equal to
(1) $\frac{9}{2^{9}}$
(2) $\frac{91}{2^{13}}$
(3) $\frac{7}{2^{7}}$
(4) $\frac{11}{2^{12}}$
Show Answer
Answer: (2)
Solution:
$2 \tan ^{2} \theta-5 \sec \theta-1=0$
$$ \begin{aligned} & \Rightarrow \quad 2\left(\sec ^{2} \theta-1\right)-5 \sec \theta-1=0 \\ & \Rightarrow \quad 2 \sec ^{2} \theta-5 \sec \theta-3=0 \\ & \Rightarrow \quad 2 \sec ^{2} \theta-6 \sec \theta+\sec \theta-3=0 \\ & \Rightarrow \quad(2 \sec \theta+1)(\sec \theta-3)=0 \\ & \quad \sec \theta=3 \\ & \Rightarrow \quad \cos \theta=1 / 3 \end{aligned} $$
2 solutions in $[0,2 \pi]$
2 solutions in $[2 \pi, 4 \pi]$
2 solutions in $[4 \pi, 6 \pi]$
1 solution in $\left[6 \pi, \frac{13 \pi}{2}\right]$
$\Rightarrow n=13$
$\sum _{K=1}^{13} \frac{K}{2^{13}}$
$\Rightarrow \frac{1}{2^{13}}(1+2$..
$=\left(\frac{13 \cdot 14}{2}\right) \cdot \frac{1}{2^{13}}$
$=\frac{13 \cdot 7}{2^{13}}=\frac{91}{2^{13}}$
BETA
