If has exactly 7 solutions in the interval for the least value of then is equal to

Question

If $2 \tan^{2} θ - 5 \sec θ = 1$ has exactly 7 solutions in the interval $[0, \frac{n π}{2}],$ for the least value of $n \in N,$ then $\sum_{k = 1}^{n} \frac{k}{2^{k}}$ is equal to [2024]

Smart Achievers · Accepted Answer

(3)

Given, $2 \tan^{2} θ - 5 \sec θ = 1$

$\Rightarrow 2 (\sec^{2} θ - 1) - 5 \sec θ = 1$

$\Rightarrow 2 \sec^{2} θ - 5 \sec θ - 3 = 0$

$\Rightarrow (2 \sec θ + 1) (\sec θ - 3) = 0 \Rightarrow \sec θ = - \frac{1}{2}, 3$

$\Rightarrow \cos θ = - 2, \frac{1}{3}$ $(\cos θ \in [- 1, 1])$

$The least value of n \in N, for which exactly 7 solutions are possible is 13 .$

So, $\sum_{k = 1}^{13} \frac{k}{2^{k}} = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots + \frac{13}{2^{13}}$

Let $S = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots + \frac{13}{2^{13}}$

$\frac{S}{2} = \frac{1}{2^{2}} + \frac{2}{2^{3}} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}$

$\Rightarrow \frac{S}{2} = \frac{1}{2} [\frac{1 - \frac{1}{2^{13}}}{1 - \frac{1}{2}}] - \frac{13}{2^{14}} \Rightarrow S = 2 (\frac{2^{13} - 1}{2^{13}}) - \frac{13}{2^{13}}$

$\Rightarrow S = \frac{2^{14} - 15}{2^{13}}$

Solution

Q.
If $2 \tan^{2} θ - 5 \sec θ = 1$ has exactly 7 solutions in the interval $[0, \frac{n π}{2}],$ for the least value of $n \in N,$ then $\sum_{k = 1}^{n} \frac{k}{2^{k}}$ is equal to [2024]

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

2 $1 - \frac{15}{2^{13}}$

3 $\frac{1}{2^{13}} (2^{14} - 15)$

4 $\frac{1}{2^{14}} (2^{15} - 15)$

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Solution

Q. If 2tan2θ-5secθ=1 has exactly 7 solutions in the interval [0,nπ2], for the least value of n∈N, then ∑k=1nk2k is equal to [2024]

1 1215(214-14)

2 1-15213

3 1213(214-15)