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JEE Mathematics 2024 Question with Solution

If 2tan⁡2(θ)−5sec⁡(θ)=12\tan^2(\theta) - 5\sec(\theta) = 1 has exactly 77 solutions in the interval [0,nπ2]\left[0, \frac{n\pi}{2}\right] for the least value of n∈Nn \in \mathbb{N}, then Σ(k=1 to n) k2k\Sigma(k=1 \text{ to } n)\ \frac{k}{2k} is equal to:

  • A

    1215(214−14)\frac{1}{2^{15}}(2^{14} - 14)

  • B

    1214(215−15)\frac{1}{2^{14}}(2^{15} - 15)

  • C

    1−152131 - \frac{15}{2^{13}}

  • D

    1213(214−15)\frac{1}{2^{13}}(2^{14} - 15)

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: 2tan⁡2θ−5sec⁡θ=12\tan^2\theta - 5\sec\theta = 1 and the equation has exactly 77 solutions in [0,nπ2]\left[0, \frac{n\pi}{2}\right] for the least nn.

Find: ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k} and then choose the correct option.

Using tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1,

2(sec⁡2θ−1)−5sec⁡θ=12(\sec^2\theta - 1) - 5\sec\theta = 1

So,

2sec⁡2θ−5sec⁡θ−3=02\sec^2\theta - 5\sec\theta - 3 = 0

Factorizing,

(2sec⁡θ+1)(sec⁡θ−3)=0(2\sec\theta + 1)(\sec\theta - 3) = 0

Hence,

sec⁡θ=−12orsec⁡θ=3\sec\theta = -\frac{1}{2} \quad \text{or} \quad \sec\theta = 3

Now sec⁡θ=−12\sec\theta = -\frac{1}{2} gives cos⁡θ=−2\cos\theta = -2, which is not possible.

Therefore,

sec⁡θ=3⇒cos⁡θ=13\sec\theta = 3 \Rightarrow \cos\theta = \frac{1}{3}

The solutions of cos⁡θ=13\cos\theta = \frac{1}{3} occur twice in each interval of length 2π2\pi. To get exactly 77 solutions, the least endpoint must reach the seventh solution, which corresponds to

nπ2=13π2\frac{n\pi}{2} = \frac{13\pi}{2}

Thus,

n=13n = 13

Now let

S=∑k=113k2kS = \sum_{k=1}^{13} \frac{k}{2^k}

Then,

S=12+222+323+⋯+13213S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{13}{2^{13}}

Also,

S2=122+223+⋯+12213+13214\frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \cdots + \frac{12}{2^{13}} + \frac{13}{2^{14}}

Subtracting,

S−S2=12+122+123+⋯+1213−13214S - \frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{13}} - \frac{13}{2^{14}}

So,

S2=(1−1213)−13214\frac{S}{2} = \left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{14}}

Hence,

S=2(1−1213)−13213=1213(214−15)S = 2\left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{13}} = \frac{1}{2^{13}}(2^{14} - 15)

Therefore, the value of the sum is 1213(214−15)\frac{1}{2^{13}}(2^{14} - 15). The solution states the correct option is B, so the answer is B.

Series Manipulation Detail

Given: n=13n = 13 from the trigonometric part.

Find: ∑k=113k2k\sum_{k=1}^{13} \frac{k}{2^k}.

Write the sum as

S=12+222+323+⋯+13213S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{13}{2^{13}}

Now multiply by 12\frac{1}{2}:

S2=122+223+324+⋯+13214\frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \cdots + \frac{13}{2^{14}}

Subtract the second equation from the first:

S−S2=12+(222−122)+(323−223)+⋯+(13213−12213)−13214=12+122+123+⋯+1213−13214\begin{aligned} S - \frac{S}{2} &= \frac{1}{2} + \left(\frac{2}{2^2} - \frac{1}{2^2}\right) + \left(\frac{3}{2^3} - \frac{2}{2^3}\right) + \cdots + \left(\frac{13}{2^{13}} - \frac{12}{2^{13}}\right) - \frac{13}{2^{14}} \\ &= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{13}} - \frac{13}{2^{14}} \end{aligned}

Now use the geometric sum:

12+122+⋯+1213=1−1213\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{13}} = 1 - \frac{1}{2^{13}}

Hence,

S2=1−1213−13214\frac{S}{2} = 1 - \frac{1}{2^{13}} - \frac{13}{2^{14}}

Therefore,

S=1213(214−15)S = \frac{1}{2^{13}}(2^{14} - 15)

So the required value is 1213(214−15)\frac{1}{2^{13}}(2^{14} - 15).

Common mistakes

  • Taking sec⁡θ=−12\sec\theta = -\frac{1}{2} as valid. This gives cos⁡θ=−2\cos\theta = -2, which is impossible because cos⁡θ∈[−1,1]\cos\theta \in [-1,1]. Always convert back to cosine and check the valid range.

  • Assuming the least nn is obtained by counting full periods only. The question asks for exactly 77 solutions, so the interval endpoint must be placed at the seventh occurrence, not merely after a whole number of cycles.

  • Using ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k} incorrectly as a geometric series directly. This is an arithmetico-geometric series, so use the standard subtraction method or the correct closed form.

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