Aspirants preparing for JEE Advanced exam can refer to previous year question papers with answers provided by BYJU’S. Referring to last year papers help aspirants for the kind of challenges that the actual exam might throw up and help them in evaluation of performance and preparation level. JEE Advanced Question Paper 2019 Maths Paper 1 solutions are given below. Students can download to practise offline.

Paper 1 - Maths

Question 1: Let M =

$\begin{bmatrix} sin^4 \theta & -1-sin^2 \theta\\ 1 + cos^2 \theta & cos^4 \theta \end{bmatrix} = \alpha I + \beta M^{-1},$

where α = α(θ) and β = β(θ) are real number, and I is the 2x2 identity matrix. If α* is the minimum of the set {α(θ) : θ Є [0, 2π]} and β* is the minimum of the set {β(θ) : θ Є [0, 2π]} then the value of α* + β* is

a) -37/16
b) -29/16
c) -31/16
d) -17/16

Solution:

Let M =
$\begin{bmatrix} sin^4 \theta & -1-sin^2 \theta\\ 1 + cos^2 \theta & cos^4 \theta \end{bmatrix} = \alpha I + \beta M^{-1}$

M² = αM + βI

a₁₁ = sin θ – 1 – sin² θ – cos² θ – cos² θ sin² θ = β + α sin⁴ θ sin⁸ θ – 2 – cos² θsin² θ

= β + α sin⁴ θ

a₂₁ = sin⁴ θ + cos² θ sin⁴ θ + cos⁴ θ + cos⁶ θ = α (1 + cos² θ)

(1 + cos² θ) α = sin⁴ θ (1 + cos² θ) + cos⁴ θ (1 + cos² θ)

α = sin⁴ θ + cos⁴ θ = 1 – 2 sin² θ cos² θ = 1 – ((sin² θ)/2)

α_min = ½

β = sin⁸ θ – 2 – cos² θsin² θ - sin⁸ θ – cos⁴ θsin⁴ θ

= -2 - (sin² 2θ)/4 - (sin⁴ 2θ)/16

β_min = -2 – 1/16 [4t² + t⁴]

= -2 – 1/16 [t² + 2]² + ¼

= -7/4 – 1/16(9) = -37/16

Therefore, α + β = -37/16 + 1/2 = -29/16

Question 2: A line y = mx + 1 intersects the circle (x - 3)² + (y + 2)² = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate -3/5, then which one of the following options is correct?

a) 6 ≤ m < 8
b) 2 ≤ m < 4
c) 4 ≤ m < 6
d) -3 ≤ m < -1

Solution:

y = mx + 1

(x - 3)² + ((mx +1) + 2)² = 25

=> x² (1+ m² ) + 6(m -1) x - 7 = 0

$\frac{-3}{5} = \frac{\alpha + \beta}{2} = \frac{-6(m-1)}{2(1+m^2}$

1 + m² = 5m – 5

Or m = 2, 3

Question 3: Let S be the set of all complex numbers z satisfying |z - 2 + i| ≥ √5. If the complex number z₀ is such that 1/|z₀ - 1| is the maximum of the set

$\begin{Bmatrix} \frac{1}{|z-1|}; z \in S \end{Bmatrix}$

then the principle argument of

$\frac{4-z_0 - \bar{z_0}}{z_0- \bar{z_0} +2i}$

a) π/4
b) -π/2
c) 3π/4
d) π/2

Solution:

|z - 2 + i| ≥ √5

P is along
$\bar {AC}$
but at √5 distance from C.

$JEE Advanced Question Paper 2019 Math Paper 1$

Question 4: The area of the region {(x, y) : xy ≤ 8, 1 ≤ y ≤ x²} is
JEE Advanced Question Papers 2019 Maths Paper 1

Solution:

{(x, y) : xy ≤ 8, 1 ≤ y ≤ x²}

8/x = x²

$\int_1^2 (x^2 - 1)dx + \int_2^8(\frac{8}{x} -1)dx$

= 16 log_e 2 – 14/3

Question 5: There are three bags B₁, B₂ and B₃. The bag B₁ contains 5 red and 5 green balls, B₂ contains 3 red and 5 green balls, and B₃ contains 5 red and 3 green balls, Bags B₁, B₂ and B₃ have probabilities 3/10, 3/10 and 4/10 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

a) Probability that the selected bag is B₃ and the chosen ball is green equals 3/10
b) Probability that the chosen ball is green equals 39/80
c) Probability that the chosen ball is green, given that the selected bag is B₃, equals 3/8
d) Probability that the selected bag is B₃, given that the chosen balls is green, equals 5/13

Solution:

$JEE Advanced Question Paper 2019 Paper 1 Math$

Question 6: Define the collections E₁, E₂, E₃ ,..... of ellipses and {R₁, R₂, R₃, …..} of rectangles as follows:

$E_1 = \frac{x^2}{9}+\frac{y^2}{4} = 1$

R₁: Rectangle of largest area, with sides parallel to the axes, inscribed in E₁;

E_n: Ellipse

$\frac{x^2}{a_n^2}+\frac{y^2}{b_n^2} = 1$

of largest area inscribed in R_n-1, n > 1; R_n: Rectangle of largest area, with sides parallel to the axes, inscribed in E_n, n > 1.

Then which of the following options is/are correct?

a) The eccentricities of E₁₈ and E₁₉ are NOT equal.
b) The distance of a focus from the centre in E₉ is √5/32
c) The length of latus rectum of E₉ is 1/6
d)
$\sum_{n=1}^N$
(area of R_n) < 24, for each positive integer N

Solution:

A = 6 cosθ . 4 sin θ = 12 sin 2θ -> max

θ = π/4

E₂ = a₂ = 3 cos(π/4) = 3/√2

b₂ = 2 sin θ = √2

$r = \frac{1}{\sqrt 2}, a_n = \frac{3}{(\sqrt{2})^{n-1}},b_n = \frac{2}{(\sqrt{2})^{n-1}}$

$JEE Advanced Math Question Paper 2019 Paper 1$

So, e₁₈ = 1 – b₂/a₂ = √5/3

e of all ellipses -> same

Difference of f from conic in equation a_qe

$E_a = \frac{2b_n^2}{a_n} = \frac{3}{(2^8)} \times \frac{\sqrt 5}{3}=\frac{\sqrt 5}{16}$

Length of latus rectum = [2x4]/4(√2) = 1/6

Question 7: Let

$M = \begin{bmatrix} 0 &1 & a \\ 1& 2 &3 \\ 3 &b &1 \end{bmatrix}$

and adj M =

$\begin{bmatrix} -1 &1 & -1 \\ 8& -6 &2 \\ -5 &3 &-1 \end{bmatrix}$

where a and b are real numbers. Which of the following options is/are correct?

a) a + b = 3
b) det(adj M²) = 81
c) (adj M)^-1 + adj M^-1 = -M
d) If
$M \begin{bmatrix} \alpha\\ \beta\\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2\\ 3 \end{bmatrix}$
, then α - β + γ = 3.

Solution:

$M = \begin{bmatrix} 0 &1 & a \\ 1& 2 &3 \\ 3 &b &1 \end{bmatrix}$
and adj M =
$\begin{bmatrix} -1 &1 & -1 \\ 8& -6 &2 \\ -5 &3 &-1 \end{bmatrix}$

(a)

adj M =
$\begin{bmatrix} 2-3b &ab-1 & -1 \\ 8& -6 &2 \\ b-6 &3 &-1 \end{bmatrix}$

b – 6 = -5

=>b = 1

and ab – 1 = 1

or a = 2

Now,
$M = \begin{bmatrix} 0 &1 & 2 \\ 1& 2 &3 \\ 3 &1 &1 \end{bmatrix}$

|M| = -2

And a + b = 2

(b) |adj M²|= |M²|² = |M|⁴ = 16

(c) (adj M)^-1 + adj M^-1 = 2 (adj M)^-1

= 2 M^-1 M

= 2 x (-1/2) x M

= -M

(d)
$M\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$

This implies,

$\begin{bmatrix} 0 &1 &2 \\ 1 &2 &3 \\ 3 & 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$

β + 2γ = 1

α + 2β + 3γ = 2

3α + β + γ = 1

α - β + γ = 3

Solving all the equations, we have α = 1, β = -1 and γ = 1

Question 8: Maths JEE Advanced Question Paper 2019 Paper 1

Then which of the following options is/are correct?

a) f’ has a local maximum at x = 1
b) f is onto
c) f is increasing on (-∞, 0)
d) f’ is NOT differentiable at x = 1

Solution:

$Math JEE Advanced Question Paper 2019 Paper 1$

Range: -∞, +1

Therefore, Not monotonic.

f(x) = x² – x + 1

f’(x) = 2x - 1

Max at x = 0 and 1

x ≥ 3

f(3) = 1 log 1 – 3 + 10/3 = 1/3

f(∞) -> ∞

$f'(x) = \left\{\begin{matrix} 2x-1 & ; 0 \leq x <1\\ 2x^2-8x+7 & ;1 \leq x<3 \end{matrix}\right.$

Loc. Max at x = 1

Question 9: Let α and β be the roots of x² – x – 1 = 0, with α > β. For all positive integers n, define

$a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, n \geq 1$

b₁ = 1 and b_n = a_n-1 + a_n+1, n ≥ 2.

Then which of the following options is/are correct?

a)
$a_1 + a_2 + a_3 +....+a_n = a_{n+2} - 1$
for all n ≥ 1
b)
$\sum_{n=1}^\infty \frac{a_n}{10^n} = \frac{10}{89}$
c)
$\sum_{n=1}^\infty \frac{b_n}{10^n} = \frac{8}{89}$
d)
$b_n = \alpha^n + \beta^n$
for all n ≥ 1.

Solution:

x² – x – 1 = 0

α = [1+ √5]/2 and β = [1- √5]/2

$a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$

As, b₁ = 1 and b_n = a_n-1 + a_n+1, n ≥ 2

a₁ + a₂ + …+ a_n =
$\sum_{n=1}^\infty \frac{a_n}{10^n}$

=> a_n + a_n+1 = a_n+2

=>
$\frac{\sum (\frac{\alpha}{10})^n - \sum(\frac{\beta}{10})^n}{\alpha - \beta}$

$\sum_{r=1}^n a_r= a_{n+2} - a_2$

= a_n+2 – [α² - β²]/[α -β]

= a_n+2 – [α + β]

= a_n+2 – 1

$Solved JEE Advanced Question Paper 2019 Paper 1 Math$

Question 10: Let denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to at a point P intersect the y-axis at Y_P. If PY_P has length 1 for each point P on , then which of the following is options is/are correct?

a)
$y = log_e(\frac{1+\sqrt{1-x^2}}{x}) - \sqrt{1-x^2}$
b) xy' - √[1-x²] = 0
c)
$y = log_e(\frac{1+\sqrt{1-x^2}}{x}) + \sqrt{1-x^2}$
d) xy' + √[1-x²] = 0

Solution:

y - y₁ = m(x – x₁)

y_p - y₁ = -mx₁

=> 1 = x₁² + m² x₁²

1 = x²[1 + (dy/dx)²]

Or (dy/dx)² = 1/x² – 1

dy/dx = ± √[1-x²]/x

$y = \pm \int \frac{\sqrt{1-x^2}}{x} \ dx$

Here x = sinθ, dx = cos θ dθ

$Solved JEE Advanced Math Question Paper 2019 Paper 1$

=>
$y = log_e(\frac{1+\sqrt{1-x^2}}{x}) - \sqrt{1-x^2}$

And the correct differential equation will be xy' + √[1-x²] = 0

Question 11 In a non-right-angle triangle ΔPQR, let p, q, r denote the lengths of the sides opposite to the angles at P, Q, R respectively. The median from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at 0. If p = √3, q = 1, and the radius of the circumcircle of the ΔPQR equals 1, then which of the following options is/are correct?

a) Area of ΔSOE = √3/12
b) Radius of incircle of ΔPQR = √3/2 [2 - √3]
c) Length of RS = √7/2
d) Length of OE = 1/6

Solution:

(sin P)/√3 = (sin Q)/1 = 1/2R = ½ ; [Given R = 1]

∠P = π/3 or 2π/3

∠Q = π/6 or 5π/6

p>q=> ∠P = ∠Q

If ∠P = π/3 and ∠Q = π/6 => ∠R = π/2

Therefore, ∠P = 2π/3 and ∠Q = ∠R = π/6

Question 12: Let L₁ and L₂ denotes the lines
Answers for JEE Advanced Maths 2019 Question Paper 1

respectively. If L₃ is a line which is perpendicular to both L₁ and L₂ and cuts both of them, then which of the following options describe(s) L₃?

$Answers for JEE Advanced Math Question Paper 1$

Solution:

$L_1 : \vec r = \hat i + \lambda(- \hat i + 2 \hat j + 2 \hat k)$

$L_2 : \vec r = \mu(-2 \hat i - \hat j + 2 \hat k)$

L₁ : [x-1]/-1 = [y-0]/2 = [z-0]/2

L₂ : x/2 = y/-1 = z/2

L₃ : x/a = y/b = z/c

L₃ = L₁ x L₂

Similarly,
$6 \hat i + 6 \hat j - 3 \hat k$

A on L₁ : (-λ + 1, 2λ, 2λ)

B on L₂ : (2μ, -μ, 2μ)

AB Δrs : (2μ, + λ - 1 -μ - 2λ, 2μ - 2λ) = (6, 6, -3) or (2, 2, -1)

=>
$\frac{2 \mu + \lambda - 1}{2} = \frac{- \mu - 2 \lambda}{2} = \frac{2 \mu - 2 \lambda}{-1} = k$

λ = [3k+1]/3 and μ = -4k – 2/3

Now, 2 μ - 2 λ + k = 0

Putting μ and λ values, we have

k = -2/9=> μ = 2/9 and λ = 1/9

So, A: (8/9, 2/9, 2/9) and B : (4/9, -2/9, 4/9)

Mid point is (2/3, 0, 1/3)

For equation of L₃;

vector A is the position vector of any point on L₃.

Answer: a, b, d

Question 13: If

$I = \frac{2}{\pi} \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{sin\ x})(2-cos 2x)}$

then 27I² equals _________

Solution:

27I² = 4

Question 14: Let the point B be the reflection of the point A(2, 3) with respect to the line 8x – 6y – 23 = 0. Let _A and _B be circle of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circle _A and _B such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is ________.

Solution:

ΔAPC and ΔBQC are similar.

BC/AC = 1/2 => 2(AC - AB) = AC

AC = 2AB = 10

Question 15: Let AP (a; d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d > 0. If AP(1; 3) ∩ AP (2; 5) ∩ AP (3; 7) = AP (a; d) then a + d equals_________

Solution:

I: 1, 4, 7, 10, 13, 16 ….. 32

II: 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52

III: 3, 10, 17, 24, 31, 38, 45, 52

52 <-- a + d --> LCM of 3, 5, 7 = 105

So, a + d = 157

Question 16: Let S be the sample space of all 3 × 3 matrices with entries from the set {0, 1}. Let the events E₁ and E₂ be given by

E₁ = {A ϵ S : det A = 0} and

E₂ = {A ϵ S : sum of entries of A is 7}.

If a matrix is chosen at random from S, then the conditional probability P(E₁|E₂) equals …….

Solution:

S: 2⁹

$P(\frac{E_1}{E_2}) = \frac{P(E_1 \cap E_2)}{P(E_2)}$

E₂ : Sum of entries 7, we need 7 - 1s and 2 0’s

Total E₂ = 9!/7!2! = 36

$\begin{vmatrix} 1 &1 &1 \\ 1& 1 &1 \\ 1& 0 &0 \end{vmatrix}$

Per |A| to be zero, both zeroes should be in the same row/column.

Therefore, 3 x 3 x 2 = 18 cases

$P(\frac{E_1}{E_2})$
= 18/36 = ½

Question 17: Three lines are given by

JEE Advanced Solved Maths 2019 Question Paper 1

Let the lines cut the plane x + y + z 1 at the points A, B and C respectively. If the area of the triangle ABC is Δ then the value of (6Δ)² equals __________

Solution:

1^st line: x = λ, y = 0, z = 0

x + y + z = 1 => λ = 1

A(1, 0, 0)

2^nd line: x = μ, y = μ, z = 0

2μ = 1 or μ = ½

B(1/2, ½, 0)

Parallels, C(1/3, 1/3, 1/3)

$JEE Advanced Solved Math 2019 Question Paper 1$

Question 18: Let ω ≠ 1 be a cube root of unity. Then the minimum of the set {|a + bω + cω²|² ; Where, a, b, c distinct non-zero integers }equals ________

Solution:

| a + bω + cω²|²

= (a + bω + cω²) (a + cω + bω²)

= (a² + b² + c² - ab – bc – ca)

[because, conjugate of (a + bω + cω²) = (a + cω + bω²)]

= > (1/2)[(a-b)² + (b-c)² + (c-a)²] = ½(1+1+4) = 3

[For minimum value a = 1, b = 2 and c = 3]

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Video Lessons - Paper 1 Maths

JEE Advanced 2019 Maths Paper 1 Solutions

JEE Advanced 2019 Maths Paper 1 Question and Solutions