JEE Main 2025 Jan 28 Shift 1, Mathematics Q22: De Moivre's Theorem And Its Applications

JEE Main2025Jan 28, Shift 1Mathematics

If $α = 1 + r = 1 \sum 6 (- 3)^{r - 1}^{12} C_{2 r - 1}$ , then the distance of the point $(12, 3)$ from the line $α x - 3 y + 1 = 0$ is _______.

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Solution

Rewrite the sum by recognising it as the imaginary part of a binomial expansion. Note $(- 3)^{r - 1} = \frac{( 3 i ) ^{2 r - 1}}{3 i} \cdot (- 1)$ wisely; we use the cleaner approach below.

Consider $(1 + 3 i)^{12}$ and $(1 - 3 i)^{12}$ .

$(1 + 3 i)^{12} - (1 - 3 i)^{12} = 2 \sum_{k odd} (k 12) (3 i)^{k} = 23 i \sum_{r = 1}^{6} (2 r - 1 12) (3 i)^{2 r - 2}$

$(3 i)^{2 r - 2} = (- 3)^{r - 1}$ . So $(1 + 3 i)^{12} - (1 - 3 i)^{12} = 23 i \sum_{r = 1}^{6} (- 3)^{r - 1} (2 r - 1 12)$ .

Also $1 + 3 i = 2 e^{iπ /3}$ and $1 - 3 i = 2 e^{- iπ /3}$ , so $(1 + 3 i)^{12} = 2^{12} e^{i 4 π} = 4096$ , similarly $(1 - 3 i)^{12} = 4096$ . The difference is 0.

$23 i \sum_{r = 1}^{6} (- 3)^{r - 1} (2 r - 1 12) = 0 \Rightarrow \sum_{r = 1}^{6} (- 3)^{r - 1} (2 r - 1 12) = 0$ .

So $α = 1 + 0 = 1$ .

Distance from $(12, 3)$ to $x - 3 y + 1 = 0$ : $\frac{∣12 - 3 \cdot 3 + 1∣}{1 + 3} = \frac{∣12 - 3 + 1∣}{2} = \frac{10}{2} = 5$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q22: De Moivre's Theorem And Its Applications

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