Let x = x(y) be the solution of the differential equation y² dx + (x – 1/y) dy = 0

Let x = x(y) be the solution of the differential equation y² dx + (x – 1/y) dy = 0

Question 12 : Let x = x(y) be the solution of the differential equation y^2 dx + (x - \dfrac{1}{y}) dy = 0. x(1) = 1 then x(1/2) is :

(i) 1/2 + e         (ii) 3/2 + e

(iii) 3 – e            (iv) 3 + e

Answer:  (iii) 3 – e

Explanation:- The given differential equation

y^2 dx + (x - \dfrac{1}{y}) dy = 0

y^2 dx =  -(x - \dfrac{1}{y}) dy

\dfrac{dx}{dy} = - \dfrac{x}{y^2} + \dfrac{1}{y^3}

\dfrac{dx}{dy} + \dfrac{x}{y^2} = \dfrac{1}{y^3}

Compare with \dfrac{dx}{dy} + Px = Q

P = 1/y² and Q = 1/y³

I.F. = e^{\int \frac{1}{y^2}} dy

= e^{-\frac{1}{y}}

Solution of the given differential equation

x. (I.F.) = \int Q. (I.F.) dy + C

x . e^{-\frac{1}{y}} = \int e^{-\frac{1}{y}}. \dfrac{1}{y^3} dy .  .  . (i)

Put -1/y = t

⇒ y = -1/t

\dfrac{1}{y^2} dy = dt

x . e^{-\frac{1}{y}} = - \int e^{t}. t dy

x . e^{-\frac{1}{y}} = - t.e^{t} + e^t + C

x . e^{-\frac{1}{y}} = \frac{1}{y}.e^{ -\frac{1}{y}} + e^{ -\frac{1}{y}} + C

x = 1 and y = 1

\dfrac{1}{e} = \dfrac{1}{e} + \dfrac{1}{e} + C

⇒ C = -1/e

Put y = 1/2

\dfrac{x}{e^2} = \dfrac{2}{e^2} + \dfrac{1}{e^2} - \dfrac{1}{e}

⇒ x = 3 -e

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 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:- See full answer

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

Question 11: From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ‘M’, is :

(1) 14950              (2) 6084

(3) 4356                (4) 5148                  [JEE 22 JAN 2025]

Answer : See full Answer

Question 13: Let the parabola y = x³ + px – 3 meet the coordinate axes at the points P, Q and R. If the circle C with centre at (-1, -1) passes through the points P, Q and R, then the area of ΔPQR is :

(i)  4                 (ii)  6

(iii)  7               (iv)  5

Answer :See full Answer