If 2 tan^2 ⁡ -5 sec ⁡ =1 has exactly 7 solutions in the interval (0 , nπ2 ), for the least value of n ∈N then _k=1^n 0.05em k2^k is equal to :

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Correct answer: D. (d) 2 tan^2 ⁡ -5 sec ⁡ -1=0 ⇒2 sec^2 ⁡ -5 sec ⁡ -3=0 ⇒ (2 sec ⁡ +1 ) ( sec ⁡ -3 )=0 ⇒ sec ⁡ =- 12 ,3 ⇒ cos ⁡ =-2 , 13 ⇒ cos ⁡ = 13 For 7 solutions n=13 So , _k=1^13 0.05em k2^k=S ( say )S= 12+ 22^2+ 32^3+ …+ 132^13 12 S= 12^2+ 12^3+ …+ 122^13+ 132^14 ⇒ S2= 12 1- 12^131- 12- 132^14 ⇒S=2 ( 2^13-12^13 )- 132^13

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

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