JEE Advanced 2019 Maths Paper 2

Then the possible value(s) of a is/are:

a) 8
b) –9
c) –6
d) 7

Answer: a, b

Question 3: Three lines

are given. For which point(s) Q and L₂ can we find a point P on L₁ and a point R on L₃ so that P, Q and R are collinear?

a)

Then which of the following options is/are correct?

a) f(x) = x|x| has PROPERTY 2
b) F(x) = x^2/3 has PROPERTY 1
c) f(x) = sin x has PROPERTY 2
d) f(x) = |x| has PROPERTY 1

Answer: b, d

Assuming cos^-1x takes value in [0, π], which of the following options is/are correct?

a) If α = tan (cos^-1f(6)), then α² + 2α – 1 = 0
b) lim_(n-> ∞) f(n) = 1/2
c) f(4) = √3/2
d) sin (7 cos^-1f(5)) = 0

(a) α = tan(cos^-1 f(6)) = tan(cos^-1(cos π/8) = tan (π/8)

α² + 2 α – 1 = tan² (π/8) + 2 tan (π/8) – 1

tan 2(π/8) = [2 tan (π/8)]/[1- tan²(π/8)]

This implies, 1 = 2α/[1- α²]

α² + 2 α – 1 = 0

Option (a) is correct.

(b)

Option (b) is not correct.

(c) f(4) = cos(π/6) = √3/2

Option (c) is correct.

(d) sin(7 cos^-1 f(5)) = sin(7 cos^-1 (cos π/7)) = sin(7 x π/7) = 0

Option (d) is correct.

Answer: a, c, d

Where P_K^T denotes the transpose of the matrix P_K. Then which of the following options is/are correct?

a) X – 30I is an invertible matrix
b) The sum of diagonal entries of X is 18
c) If

And R = PQP^-1. Then which of the following options is/are correct?

a) For x = 1, there exists a unit vector

Let x₁ < x₂ < x₃ <…..< x_n < …. be all the points of local maximum of f.

and y₁ < y₂ < y₃ <…..< y_n < …. be all the points of local minimum of f.

Then which of the following options is/are correct?

a) |x_n – y_n| > 1 ;for every n
b) x₁ < y₁
c) x_n Є (2n, 2n + 1/2) ;for every n
d) x_n+1 – x_n > 2 ;for every n

*for students, Draw a graph and conclude.

holds for some positive integer n. Then

Let f(x) = sin(π cosx) and g(x) = cos(2π sinx) be two functions defined for x > 0. Define the following sets whose element are written in the increasing order:

X = {x: f(x) = 0}, Y = {x: f'(x) = 0}

Z = {x: g(x) = 0}, W = {x: g'(x) = 0}

List –I contains the sets X,Y,Z and W. List – II contains some information regarding these sets.

Which of the following is the only correct combination?

a) (II), (R), (S)
b) (I), (P), (R)
c) (II), (Q), (T)
d) (I), (Q), (U)

Note: For answer, check next solution.

Let f(x) = sin(π cosx) and g(x) = cos(2π sinx) be two functions defined for x > 0. Define the following sets whose element are written in the increasing order:

X = {x: f(x) = 0}, Y = {x: f'(x) = 0}

Z = {x: g(x) = 0}, W = {x: g'(x) = 0}

List –I contains the sets X,Y,Z and W. List – II contains some information regarding these sets.

Which of the following is the only correct combination?

a) (IV), (Q), (T)
b) (IV), (P), (R), (S)
c) (III), (R), (U)
d) (III), (P), (Q), (U)

f(x) = 0

=> sin(π cos x) = 0

This implies, π cos x = nπ

Cos x = n

=>Cos x = -1, 0, 1

x = {nπ , (2nπ) π /2}

=> x = {nπ/2, n Є I}

f’(x) = 0

=> cos(π cos x)(- π sin x) = 0

=> π cos x = (2n + 1) π/2 or x = nπ

=> cos x = n + (1/2) or x = nπ

=> cos x = ±(1/2) or x = nπ

Therefore, y = {2nπ ± π/3, 2nπ ± 2π/3, n π }

g(x) = 0 => cos (2 π sin x) = 0

=> 2 π sin x = (2n + 1) π/2

=> sin x = (2n+1)/4 = ±(1/4), ±(3/4)

z = {nπ ± sin^-1(1/4), nπ ± sin^-1(3/4), n Є I}

g’(x) = 0 => -sin (2 π sin x) 2 π cos x = 0

=> 2 π sin x = n π or x = (2n + 1) π/2

=> sin x = n/2 = 0, ±(1/2), ±1 or x = (2n + 1) π/2

And, w = {nπ, (2n + 1) π/2, nπ ± π/6, n Є I}

Let the circles C₁ : x² + y² = 9 and C₂ : (x – 3) ² + (y – 4) ² = 16, intersect at the points X and Y. Suppose that another circle C₃ : (x – h) ² + (y – k) ² = r² satisfies the following conditions:

(i) centre of C₃ is collinear with the centres of C₁ and C₂

(ii) C₁ and C₂ both lie inside C₃, and

(iii) C₃ touches C₁ at M and C₂ at N

Let the line through X and Y intersect C₃ at Z and W, and let a common tangent of C₁ and C₃ be a tangent to the parabola x² = 8αy.

There are some expressions given in the List – I whose values are given in List – II below:

Which of the following is the only INCORRECT combination?

a) (IV), (S)
b) (IV), (U)
c) (III), (R)
d) (I), (P)

(i) C₁ , C₂ and C₃ are collinear.

3k = 4h….(a)

Diameter of C₃ = 12, so radius = 6 …(b)

C₃ touches C₁

C₁(0, 0) and C₃(h, k)

|C₁C₃| = |r – 3|

So, h² + k² = 9…(c)

From (a) and (c)

h = ±9/5 and k = ±12/5

=> 2h + k = 6

(ii) Equation of line ZW

C₁ = C₂

=> 3x + 4y = 9

=> Distance of ZW from (0, 0)

(ii) Area of ΔmZN = (1/2). Nm. (1/2 . ZW) = 72√6/5

Area of ΔZmW = (1/2). ZW . (om + op) = (1/2) . 24√6/5 . (3 + 9/5) = 288√6/25

Therefore, [Area of ΔmZN]/ [Area of ΔZmW] = 5/4

(iv) Slope of tangent to C₁ at m = -3/4

Equation of Tangent, y = mx – 2√(1+m²)

y = -3x/4 – 3√[1+9/16] = -3x/4 – 15/4

x = -4y/3 – 5 …(i)

Tangent to x² = 4 (2α) y is x = m’y = 2α/m’ …(ii)

On comparing (i) and (ii), we get

m’ = -4/3 and 2α/m’ = -5 => α = 10/3

Let the circles C₁ : x² + y² = 9 and C₂ : (x – 3) ² + (y – 4) ² = 16, intersect at the points X and Y. Suppose that another circle C₃ : (x – h) ² + (y – k) ² = r² satisfies the following conditions:

(i) centre of C₃ is collinear with the centres of C₁ and C₂

(ii) C₁ and C₂ both lie inside C₃, and

(iii) C₃ touches C₁ at M and C₂ at N

Let the line through X and Y intersect C₃ at Z and W, and let a common tangent of C₁ and C₃ be a tangent to the parabola x² = 8αy.

There are some expressions given in the List – I whose values are given in List – II below:

Which of the following is the only INCORRECT combination?

a) (II), (T)
b) (I), (S)
c) (I), (U)
d) (II), (Q)

*Note: Refer below solution.