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Mathematics · Complex Numbers

JEE Main 2025 — 28 January, Morning Shift — Question 22

If α=1+∑r=16(−3)r−112C2r−1\alpha=1+\sum_{\mathrm{r}=1}^{6}(-3)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1}, then the distance of the point (12,3)(12, \sqrt{3}) form the line

αx−3y+1=0\alpha x-\sqrt{3} y+1=0 is

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

α=1+∑r=16(−1)r−112C2r−13r−1\alpha=1+\sum_{\mathrm{r}=1}^{6}(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1}

α=1+∑r=1612C2r−1(3i)2r−13ii= iota, let 3i=xα=1+13i(12C1x+12C3x3+….12C11x11)=1+13i((1+3i)12−(1−3i)122)=1+13i((−2w2)12−(2w)122)=1\begin{aligned} \alpha & =1+\sum_{\mathrm{r}=1}^{6}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{r}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=\text { iota, let } \sqrt{3} \mathrm{i}=\mathrm{x} \alpha \\& =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_{1} \mathrm{x}+{ }^{12} \mathrm{C}_{3} \mathrm{x}^{3}+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}\right) \\& =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}\right) \\& =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \mathrm{w}^{2}\right)^{12}-(2 \mathrm{w})^{12}}{2}\right)=1 \end{aligned}

so distance of (12,3)(12, \sqrt{3}) from x−3y+1=0x-\sqrt{3} y+1=0 is

12−3+12=5\frac{12-3+1}{2}=5

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Complex Numbers
Topic
Demoivre's Theorem and Roots of Unity

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