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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 ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k} 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

Find: The least value of nn such that the equation has exactly 77 solutions in [0,nπ2]\left[0, \frac{n\pi}{2}\right], and then evaluate ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k}.

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−2−5sec⁡θ=12\sec^2\theta - 2 - 5\sec\theta = 1 2sec⁡2θ−5sec⁡θ−3=02\sec^2\theta - 5\sec\theta - 3 = 0

Let x=sec⁡θx = \sec\theta. Then

2x2−5x−3=02x^2 - 5x - 3 = 0

Factoring,

(2x+1)(x−3)=0(2x + 1)(x - 3) = 0

Hence,

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

Since sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta},

cos⁡θ=−2orcos⁡θ=13\cos\theta = -2 \quad \text{or} \quad \cos\theta = \frac{1}{3}

But cos⁡θ=−2\cos\theta = -2 is not possible. Therefore,

cos⁡θ=13\cos\theta = \frac{1}{3}

For cos⁡θ=13\cos\theta = \frac{1}{3}, there are two solutions in each interval of length 2π2\pi. To get exactly 77 solutions, the least endpoint must reach the seventh occurrence, which gives n=13n = 13 as concluded in the solution working.

Now evaluate

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

Write

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

Then

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}}

Therefore,

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 correct option is B.

Note: The extracted option text shows this value under option D, while the solution explicitly states The Correct Option is B and concludes with 1213(214−15)\frac{1}{2^{13}}(2^{14} - 15).

Solution from extracted working

Given: 2tan⁡2θ−5sec⁡θ−1=02\tan^2\theta - 5\sec\theta - 1 = 0

Find: ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k} for the required least nn.

Rewriting,

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

Then,

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

Thus,

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

Hence,

cos⁡θ=−2,cos⁡θ=13\cos\theta = -2, \quad \cos\theta = \frac{1}{3}

Since cos⁡θ=−2\cos\theta = -2 is impossible,

cos⁡θ=13\cos\theta = \frac{1}{3}

From the extracted second approach, the series is taken as

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

Now,

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

and

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

Subtracting gives

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

Thus,

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

So,

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

Therefore, the answer concluded by the solution is 1213(214−15)\frac{1}{2^{13}}(2^{14} - 15), and the page marks the correct option as B.

Common mistakes

  • Rejecting sec⁡θ=−12\sec\theta = -\frac{1}{2} for the wrong reason. The correct reason is that it implies cos⁡θ=−2\cos\theta = -2, which is impossible because cos⁡θ\cos\theta lies in [−1,1][-1,1]. Always convert back to cosine and check the valid range.

  • Counting the number of solutions of cos⁡θ=13\cos\theta = \frac{1}{3} incorrectly over successive intervals. This equation gives repeated solutions periodically, so the interval length must be tracked carefully rather than assuming a fixed count without verification.

  • Using the finite-series formula incorrectly for ∑k=1nk2k\sum_{k=1}^{n} \frac{k}{2^k}. A common error is to miss the last correction term after subtracting S2\frac{S}{2} from SS. Align the shifted series term-by-term before subtraction.

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