JEE Main 202427 Jan, Evening shiftMathematicsMediumTrigonometry118× since 2002If 2tan2θ−5secθ=12 \tan ^2 \theta-5 \sec \theta=12tan2θ−5secθ=1 has exactly 7 solutions in the interval [0,nπ2]\left[0, \frac{n \pi}{2}\right][0,2nπ], for the least value of n∈Nn \in \mathbf{N}n∈N, then \sum_\limits{k=1}^n \frac{k}{2^k} is equal to:AOption A. 1214(215−15)\frac{1}{2^{14}}\left(2^{15}-15\right)2141(215−15)BOption B. 1−152131-\frac{15}{2^{13}}1−21315COption C. 1215(214−14)\frac{1}{2^{15}}\left(2^{14}-14\right)2151(214−14)DOption D. 1213(214−15)\frac{1}{2^{13}}\left(2^{14}-15\right)2131(214−15)CheckSkip · show solution
