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Paper 2 - Maths

Question 1. Let f: R -> R be given by f(x) = (x - 1) (x - 2) (x - 5). Define F(x) =

$\int_0^x f(t) dt, x > 0$

. Then which of the following options is/are correct?

a) F has a local minimum at x = 1
b) F has a local maximum at x = 2
c) F(x) ≠ 0 for all x ϵ (0, 5)
d) F has two local maxima and one local minimum in (0, ∞)

Solution:

F(x) = (x - 1) (x - 2) (x - 5)

At x = 1 and x = 5, F'(x) changes from - to +

Therefore, F(x) has two local minima points at x = 1 and x = 5.

F(x) has one local maxima point at x = 2.

Answer: a,b,c

Question 2. $JEE Advanced Question Paper 2019 Math Paper 2$

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

a) 8
b) –9
c) –6
d) 7

Solution:

Answer: a, b

Question 3: Three lines $JEE Advanced 2019 Question Paper Math Paper 2$

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)
$\hat k + \hat j$
b)
$\hat k$
c)
$\hat k + (1/2) \hat j$
d)
$\hat k – (1/2) \hat j$

Solution:

P(λ, 0, 0), Q(0, μ, 1), R(1, 1, r)

Given:
$\vec {PQ} = k . \vec {PR}$

λ/[ λ-1] = - μ/-1 = -1/-r

μ cannot take the values 0 and 1.

Answer: c, d

Question 4: Let F: R -> R be a function. We say that f has
JEE Advanced 2019 Question Paper Maths Paper 2 Question 4

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

Solution:

Answer: b, d

Question 5: For non-negative integers n, let
JEE Advanced 2019 Question Paper Maths Paper 2 Question 5

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

Solution:

(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

Question 6: JEE Advanced 2019 Question Paper Maths Paper 2 Question 6

Where P_K^Tdenotes 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
$X \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = \alpha \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$
, then α = 30
d) X is a symmetric matrix

Solution:

P₁ = P₁^T = P₁^-1

P₂ = P₂^T = P₂^-1

…….

P₆ = P₆^T = P₆^-1

And Let
$A = \begin{bmatrix} 2 &1 &3 \\ 1 & 0& 2\\ 3& 2 &1 \end{bmatrix}$

Here A^T = A -> A is symmetric matrix

Now, X^T = (P₁AP₁^T + ……+ P₆AP₆^T)^T

= P₁AP₁^T + ……+ P₆AP₆^T

= X

Therefore, X is symmetric matrix.

Again,

Let B =
$\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

XB = P₁AP₁^TB + P₂AP₂^TB +……..+ P₆AP₆^TB

= P₁AB + P₂AB+ ……+ P₆AB

= (P₁ + P₂+ ……+ P₆)
$\begin{bmatrix} 6\\ 3\\ 6 \end{bmatrix}$

=
$\begin{bmatrix} 30\\ 30\\ 30 \end{bmatrix}$
= 30 B

=> α = 30

Since
$X \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = 30 \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

=>(X – 30I)B = 0 has a nontrivial solution B =
$\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

=>X – 30I = 0

X = P₁AP₁^T + …..+ P₆AP₆^T

Trace(X) = tr(P₁AP₁^T) + …..+ tr(P₆AP₆^T)

= (2 + 0 + 1) + ……(2 + 0 + 1) = 3 + 3 + ….6 times = 18

Answer: b, c, d

Question 7: Let x ϵ R and let
JEE Advanced 2019 Question Paper Maths Paper 2 Question 7

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

a) For x = 1, there exists a unit vector
$\alpha \hat i + \beta \hat j + \gamma \hat k \ for \ which \ R \begin{bmatrix} \alpha\\ \beta\\ \gamma \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
b) There exists a real number x such that PQ = QP
c) det R =
$det\begin{bmatrix} 2 & x & x\\ 0& 4 & 0\\ x&x & 5 \end{bmatrix} + 8$
, for all x Є R.
d) For x = 0, if
$R\begin{bmatrix} 1\\ a\\ b \end{bmatrix} = 6\begin{bmatrix} 1\\ a\\ b \end{bmatrix}$
, then a + b = 5.

Solution:

R = PQP^-1

|R| = |P||Q||P^-1|

=>det Q = 2(24) – x(0) + x(-4x) = 48 – 4x²

-4 + a + 2b/3 = 0 and -2a + 4b/3 = 0 => a = 2 and b = 3

Therefore, a + b = 5

PQ = QP

=>x + 2 + x = 2 + 2x ->No value exists.

Question 8: Let f(x) = sinπx/x² , x > 0

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

Solution:

*for students, Draw a graph and conclude.

Question 9: The value of

$sec^{-1}(1/4 \sum_{k=0}^{10} sec(7\pi/12 + k\pi/2))sec(7\pi/12 + (k+1)\pi/2))$

in the interval [-π/4, 3π/4] equals

Solution:

Question 10: Let |X| denote the number of elements in set X. Let S = {1,2,3,4,5,6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A,B) such that 1 < |B| < |A|, equals.

Solution:

The number of ordered pairs of (A, B) are

⁶C₁(⁶C₂ + ⁶C₃ + ⁶C₄ +…..+⁶C₆ )+ ⁶C₂ (⁶C₃ + ⁶C₄ + …+ ⁶C₆₎ + ⁶C₃ (⁶C₄ + ⁶C₅ +⁶C₆) + ⁶C₄ (⁶C₅ + ⁶C₆) + ⁶C₅ . ⁶C₆

= (⁶C₁ . ⁶C₂+ ⁶C₁ . ⁶C₃+…..+ ⁶C₁ . ⁶C₆)+ (⁶C₂ . ⁶C₃+ ⁶C₂ . ⁶C₄+…..+ ⁶C₂ . ⁶C₆) + (⁶C₃ . ⁶C₄+ ⁶C₃ . ⁶C₅+ ⁶C_{3 .}⁶C₆) + ⁶C_{4 .}⁶C₅+ ⁶C_4.⁶C₆ + ⁶C_{5 .}⁶C₆

= (¹²C₅ – ⁶C₁) + (¹²C₄ – ⁶C₂) + (¹²C₃ – ⁶C₃) + (¹²C₂ – ⁶C₄) + (¹²C₁ – ⁶C₅)

= (¹²C₁+ ¹²C₂+ ¹²C₃+ ¹²C₄+ ¹²C₅) – ( ⁶C₁+ ⁶C₂+ ⁶C₃+ ⁶C₄+ ⁶C₅)

= 1585 – 62 = 1523

Question 11: Five person A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is

Solution:

Maximum number of hats used of same colour are 2.

They cannot be 3 otherwise atleast 2 hats of same colour are consecutive.

Now the hats used are consider as B B G G R

Which can be selected in 3 ways.

It can be R G G B B or R R G B B or G G R R B

The number of ways of distributing blue hat (single one) in 5 persons equal to 5.

Now either position B and D are filled by green hats and C and E are filled by Red hats or B & D are filled by Red hats and C & E are filled by Green hats.

So, only 2 ways are possible.

Hence, number of ways = 3×5×2 = 30 ways.

Question 12: Suppose
JEE Advanced Question Paper Maths 2019 Paper 2 Solution

holds for some positive integer n. Then

$\sum_{k=0}^n \frac{^nC_k}{k+1}$

equals.

Solution:

Question 13: The value of the integral

$\int_0^{\frac{\pi}{2}} \frac{3 \sqrt{cos \theta}}{(\sqrt{cos \theta} + \sqrt{sin \theta})^5} \ d \theta$

equals

Solution:

$JEE Advanced Solved Question Paper Math 2019 Paper 2$

Let tan θ = t² =>sec 2θ dθ = dt

2I/3 =
$\int_0^{\infty} \frac{2tdt}{(t+1)^4}$

Hint: Split 2t/((t+1)⁴ = 1/(t+1)³ – 1/(t+1)⁴

Solving above integration, we get

I = 1/2

Question 14: Let

$\vec a = 2 \hat i + \hat j - \hat k \ and \ \vec b = \hat i + 2 \hat j + \hat k$

be two vectors. Consider a vector

$\vec c = \alpha \vec a + \beta \vec b , \alpha, \beta \in R$

. If the projection of vector c on the vector (a+b) is 3√2, then the minimum value of

$(\vec c - (\vec a \times \vec b). \vec c)$

equals

Solution:

Question 15: Answer the following by appropriately matching the lists based on the information given in the paragraph

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.

$Solved JEE Advanced Question Paper Math 2019 Paper 2$

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)

Solution:

Note: For answer, check next solution.

Question 16: Answer the following by appropriately matching the lists based on the information given in the paragraph.

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.

Solved JEE Advanced Maths 2019 Paper 2 Question Paper

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)

Solution:

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}

Question 17: Answer the following by appropriately matching the lists based on the information given in the paragraph.

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:

Solved JEE Advanced Maths 2019 Paper 2 Question Papers

Which of the following is the only INCORRECT combination?

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

Solution:

(i) C₁ , C₂ and C₃ are collinear.

$\begin{vmatrix} 0 &0 &1 \\ 3 & 4 &1 \\ h & k &1 \end{vmatrix}=0$

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

[Neglect -ve values]

=> 2h + k = 6

(ii) Equation of line ZW

C₁ = C₂

=> 3x + 4y = 9

=> Distance of ZW from (0, 0)

$|\frac{-9}{\sqrt{3^2 + 4^2}}| = \frac{9}{5}$

$Solved JEE Advanced Math 2019 Paper 2 Question Paper$

(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

Question 18: Answer the following by appropriately matching the lists based on the information given in the paragraph.

(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:

JEE Advanced Solutions for Maths 2019 Paper 2

Which of the following is the only INCORRECT combination?

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

Solution:

*Note: Refer below solution.

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