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JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Binomial Theorem

If

α = 1 + \sum_{r = 1}^{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

Answer: 5

Solution

Let

x = 3 i

. Since

x^{2} = - 3

,

(- 3)^{r - 1} = \frac{x ^{2 r - 1}}{x}

, so

r = 1 \sum 6 (- 3)^{r - 1}^{12} C_{2 r - 1} = \frac{1}{x} r = 1 \sum 6^{12} C_{2 r - 1} x^{2 r - 1} = \frac{1}{x} \cdot \frac{( 1 + x ) ^{12} - ( 1 - x ) ^{12}}{2}

Now

1 + x = 1 + 3 i = 2 (cos 6 0^{\circ} + i sin 6 0^{\circ})

, so

(1 + x)^{12} = 2^{12} (cos 72 0^{\circ} + i sin 72 0^{\circ}) = 2^{12}

.

Similarly

1 - x = 2 (cos 6 0^{\circ} - i sin 6 0^{\circ})

, so

(1 - x)^{12} = 2^{12}

as well.

The numerator

(1 + x)^{12} - (1 - x)^{12} = 0

, so the whole sum is

0

, giving

α = 1 + 0 = 1

.

The line is

x - 3 y + 1 = 0

. Distance from

(12, 3)

:

d = \frac{∣12 - 3 \cdot 3 + 1∣}{1 ^{2} + ( 3 ) ^{2}} = \frac{∣12 - 3 + 1∣}{2} = \frac{10}{2} = 5

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer

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