Let the triangle PQR be the image of the triangle with vertices

Question 3:- Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of ΔPQR is the point (α, β), then 15(α – β) is equal to :

(1) 24 (2) 19

(3) 21 (4) 22 [JEE 22 JAN 2025]

Answer:- (4) 22

Explanation:- Let ‘G’ be the centroid of triangle formed by (1, 3), (3, 1) and (2, 4)

$x = \dfrac{x_1 + x_2 + x_3}{3}= \dfrac{1 + 3 + 2}{3} = 2$

$y = \dfrac{y_1 + y_2 + y_3}{3}= \dfrac{3 + 1 + 4}{3} = 8/3$

G ≅ (2, 8/3)

Given, image of G is G’ = (α, β)

Then, $\dfrac{\alpha + 2}{2} = x$

$\dfrac{\beta + 8/3}{2} = y$

(x, y) lies on line x + 2y = 2

$\dfrac{\alpha + 2}{2} + \dfrac{\beta + 8/3}{2} = 2$

⇒ α + 2β = -10/3 . . . (i)

Since, x + 2y = 2

Slope of the equation (m) = -a/b = -1/2

Slope of perpendicular line = -1/m = 2

Euqation of line joining G and G’

G(2, 8/3) and G’ (α, β)

$\dfrac{\beta - 8/3}{\alpha - 2} = 2$

⇒ 2α – 4 = β – 8/3

⇒ 2α – β = – 4/3 . . .(ii)

Solving equation (i) and (ii)

α = -2/15 and β = -24/15

Now 15(α – β) = 15(-2/15 + 24/15)

= 22

Question 1:- The number of non-empty equivalence relations on the set {1, 2, 3} is :

(i) 6 (ii) 7

(iii) 5 (iv) 4 [JEE 22 JAN 2025]

Solution :- See full solution

Question 2:- Let f : R → R be a twice differentiable function such that f(x + y) = f(x) f(y) for all x, y ∈ R. If f'(0) = 4a and f satisfies f”(x) – 3a f'(x) – f(x) = 0, a > 0, then the area of the region

R = {(x, y) | 0 ≤ y ≤ f(ax), 0 ≤ x ≤ 2} is :

(i) $e^2 - 1$ (ii) $e^4 + 1$

(iii) $e^4 - 1$ (iv) $e^2 + 1$ [JEE 22 JAN 2025]

Solution:- See full solution