JEE Main 2025 Jan 28 Shift 1, Mathematics Q17: Combinations

JEE Main2025Jan 28, Shift 1Mathematics

Let $^{n} C_{r - 1} = 28$ , $^{n} C_{r} = 56$ and $^{n} C_{r + 1} = 70$ . Let $A (4 cos t, 4 sin t)$ , $B (2 sin t, - 2 cos t)$ and $C (3 r - n, r^{2} - n - 1)$ be the vertices of a triangle $A B C$ , where $t$ is a parameter. If $(3 x - 1)^{2} + (3 y)^{2} = α$ is the locus of the centroid of triangle $A B C$ , then $α$ equals :

A
$20$
B
$8$
C
$6$
D
$18$

Solution

Step 1: find $n$ and $r$ . From the ratios of consecutive binomial coefficients:

$\frac{^{n} C _{r - 1}}{^{n} C _{r}} = \frac{r}{n - r + 1} = \frac{28}{56} = \frac{1}{2} \Rightarrow 2 r = n - r + 1 \Rightarrow 3 r = n + 1$ ... (i)

$\frac{^{n} C _{r}}{^{n} C _{r + 1}} = \frac{r + 1}{n - r} = \frac{56}{70} = \frac{4}{5} \Rightarrow 5 (r + 1) = 4 (n - r) \Rightarrow 9 r = 4 n - 5$ ... (ii)

Solving (i) and (ii): $r = 3$ , $n = 8$ .

Step 2: centroid. With these values, $C = (3 r - n, r^{2} - n - 1) = (1, 0)$ .

Centroid $(x, y)$ of $A, B, C$ :

$3 x = 4 cos t + 2 sin t + 1 \Rightarrow 3 x - 1 = 4 cos t + 2 sin t$ .
$3 y = 4 sin t - 2 cos t + 0 \Rightarrow 3 y = 4 sin t - 2 cos t$ .

Compute $(3 x - 1)^{2} + (3 y)^{2}$ :

$(4 cos t + 2 sin t)^{2} + (4 sin t - 2 cos t)^{2} = 16 cos^{2} t + 16 sin^{2} t + 4 sin^{2} t + 4 cos^{2} t + (cross terms cancel) = 16 + 4 = 20$ .

$∴ α = 20$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q17: Combinations

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