Sankalp JEE Full Test-1 Question-15 Solution

Question 15. 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_{k=1}^{n} \frac{k}{2^{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 (4)

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

For 7 solutions n=13
So,

\sum_{k=1}^{13}\frac K{2^k}=S\left(say\right) S=\frac12+\frac2{2^2}+\frac3{2^3}+….+\frac{13}{2^{13}} \frac{1}{2} S=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots \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 .\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}