Binomial Theorem — JEE Main · 2025 · 28 Jan 2025 · Shift 1 Question with Solution

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 ABC, where

t

is a parameter. If

(3 x - 1)^{2} + (3 y)^{2} = λ

is the locus of the centroid of triangle ABC, then

λ

equals:

Answer: (A)

Solution

From

^{n} C_{r - 1} = 28,^{n} C_{r} = 56,^{n} C_{r + 1} = 70

\frac{^{n} C _{r - 1}}{^{n} C _{r}} = \frac{r}{n - r + 1} = \frac{28}{56} = \frac{1}{2} \Rightarrow n - r + 1 = 2 r \Rightarrow n = 3 r - 1

\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 + 5 = 4 n

Substituting

n = 3 r - 1

9 r + 5 = 12 r - 4 \Rightarrow r = 3, n = 8

. (Check:

^{8} C_{2} = 28,^{8} C_{3} = 56,^{8} C_{4} = 70

.)

So

C = (3 r - n, r^{2} - n - 1) = (1, 0)

.

Centroid of

A (4 cos t, 4 sin t), B (2 sin t, - 2 cos t), C (1, 0)

x = \frac{4 cos t + 2 sin t + 1}{3} \Rightarrow 3 x - 1 = 4 cos t + 2 sin t

y = \frac{4 sin t - 2 cos t}{3} \Rightarrow 3 y = 4 sin t - 2 cos t

(3 x - 1)^{2} + (3 y)^{2} = (4 cos t + 2 sin t)^{2} + (4 sin t - 2 cos t)^{2}

Expanding, the cross terms

16 sin t cos t

cancel, leaving

16 cos^{2} t + 4 sin^{2} t + 16 sin^{2} t + 4 cos^{2} t = 20 (cos^{2} t + sin^{2} t) = 20

λ = 20

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option