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

Question 4:- Let Z_1, Z_2 and Z_3 be three complex numbers on the circle |Z| = 1 with \arg(Z_1) = \dfrac{-\pi}{4},\arg(Z_2) = 0 and \arg(Z_3) = \dfrac{\pi}{4}. If |Z_1.\bar{Z_2} + Z_2.\bar{Z_3} + Z_3.\bar{Z_1}|^2 = \alpha + \beta\sqrt{2}, \alpha,\beta \in Z, then the value of α² + β² is :

(1) 24                (2) 41

(3) 31                (4) 29                      [JEE 22 JAN 2025]

Answer : See full Answer

Question 5:- Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of  16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2) is :

(1) 24 π²             (2) 18 π²

(3) 31 π²             (4) 22 π²                    [JEE 22 JAN 2025]

Answer :- See full answer 

Question 6:- A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ² denote the mean and variance of X, then the value of 64(μ + σ²) is :

(i)  51                 (ii)  48

(iii) 32                (iv) 64                              [JEE 22 JAN 2025]

Answer:- See full solution

Question 7:-  Let a_1, a_2, a_3 . . . be a G.P. of increasing positive terms. If a_1.a_5 = 28 and a_2 + a_4 = 29, the a_6 is equal to

(i) 628               (ii) 526

(iii) 784              (iv) 812                                    [JEE 22 JAN 2025]

Answer:- See full Answer       

Question 8:-  Let L_1: \dfrac{x - 1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} and L_2 : \dfrac{x - 2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5} be two lines. Then which of the following points lies on the line of the shortest distance between L_1 and L_2 ?

(i) (-5/3, -7, 1)            (ii) (2, 3, 1/3)

(iii) (8/3, -1, 1/3)         (iv) (14/3, -3, 22/3)                        [JEE 22 JAN 2025]

Answer:- See full solution

Question 9:- The product of all solutions of the equation e^{5(\log_e x)^2 + 3} = x^8, x > 0 is :

(i) e^{8/5}                 (ii) e^{6/5}

(iii) e^{2}                  (iv) e                              [JEE 22 JAN 2025]

Answer :- See full solution

Question 10:-  If \displaystyle \Sigma_{r= 1}^n T_r = \dfrac{(2n - 1)(2n + 1)(2n + 3)(2n + 5)}{64}, then \displaystyle \lim_{n \to \infty} {\Sigma _{r = 1}^n}\dfrac{1}{T_r} is equal to

(i) 1                     (ii) 0

(iii) 2/3                (iv) 1/3                                       [JEE 22 JAN 2025]

Answer :- See full Answer