Verified Solution • 2025–26 Session moderate

QJEEM24Ma6d4b: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6d4b Solution: If has exactly 7 solutions in the interval , for the least…

If 2tan⁡2θ−5sec⁡θ=12 \tan ^2 \theta-5 \sec \theta=1 has exactly 7 solutions in the interval [0,nπ2]\left[0, \frac{n \pi}{2}\right] , for the least value of n∈Nn \in \mathbf{N} , then ∑k=1nk2k is equal to:

Multiple Choice Options

  • A 1214(215−15)\frac{1}{2^{14}}\left(2^{15}-15\right)
  • B 1−152131-\frac{15}{2^{13}}
  • C 1215(214−14)\frac{1}{2^{15}}\left(2^{14}-14\right)
  • D 1213(214−15)\frac{1}{2^{13}}\left(2^{14}-15\right)

Step-by-Step Verified Solution

amp;2tan⁡2θ−5sec⁡θ−1=0amp;⇒2sec⁡2θ−5sec⁡θ−3=0amp;⇒(2sec⁡θ+1)(sec⁡θ−3)=0amp;⇒sec⁡θ=−12,3amp;⇒cos⁡θ=−2,13amp;⇒cos⁡θ=13\begin{aligned}& 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}\end{aligned}

For 7 solutions n=13\mathrm{n}=13

amp; So, ∑k=113k2k=S (say) amp;S=12+222+323+….+13213amp;12 S=122+123+….+12213+13214amp;⇒S2=12⋅1−12131−12−13214⇒S=2⋅(213−1213)−13213\begin{aligned}& \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \\& \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \\& \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\& \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}\end{aligned}

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