Let L1: (x - 1)/(2)=(y-2)/(3)=(z-3)/(4) and L2 : (x

Let L1: (x – 1)/(2)=(y-2)/(3)=(z-3)/(4) and L2 : (x – 2)/(3)=(y-4)/(4)=(z-5)/(5)

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:- (iv) (14/3, -3, 22/3)

Explanation:- Given Ist lines

$L_1: \dfrac{x - 1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$

Let point on the line $L_1$ is P

P(2λ + 1, 3λ + 2, 4λ + 3) on $L_1$

And the IInd line

$L_2 : \dfrac{x - 2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$

Let point on the line $L_2$ is Q

Q(3μ + 2, 4μ + 4, 5μ + 5) on $L_2$

Dr’s of PQ = (3μ – 2λ + 1, 4μ – 3λ + 2, 5μ – 4λ + 2)

Since, PQ⊥ $L_1$

Dr’s of $L_1$ (2, 3, 4)

2(3μ – 2λ + 1) + 3(4μ – 3λ + 2) + 4(5μ – 4λ + 2) = 0

38μ – 29λ + 16 = 0 . . . (i)

Again PQ⊥ $L_2$

Dr’s of $L_2$ (3, 4, 5)

3(3μ – 2λ + 1) + 4(4μ – 3λ + 2) + 5(5μ – 4λ + 2) = 0

50 μ – 38λ + 21= 0 . . . (ii)

Solving equation (i) and (ii)

λ = 1/3 ; μ = -1/6

Replacing the value of μ and λ in P and Q

P(2×1/3 + 1, 3×1/3 + 2, 4×1/3 + 3)

⇒ P(5/3, 3, 13/3)

Q(3×-1/6+ 2, 4×-1/6 + 4, 5×-1/6 + 5)

⇒ Q(3/2, 10/3, 25/6)

Line PQ

$\dfrac{x - 5/3}{1/6} = \dfrac{y-3}{-1/3} = \dfrac{z-13/3}{1/6}$

⇒ $\dfrac{x - 5/3}{1} = \dfrac{y-3}{-2} = \dfrac{z-13/3}{1}$

Points (14/3, -3, 22/3) lies on PQ

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