JEE Advanced 2019 Maths Paper 1

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 = [sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1,\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

  1. a) -37/16
  2. b) -29/16
  3. c) -31/16
  4. d) -17/16

Solution:

  1. Let M = [sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1\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)

    αmin = ½

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

    = -2 - (sin2 2θ)/4 - (sin4 2θ)/16

    βmin = -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?

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

Solution:

  1. y = mx + 1

    (x - 3)2 + ((mx +1) + 2)2 = 25

    => x2 (1+ m2 ) + 6(m -1) x - 7 = 0

    35α+β26(m1)2(1+m2\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 {1z1;zS}\begin{Bmatrix} \frac{1}{|z-1|}; z \in S \end{Bmatrix} then the principle argument of 4z0z0ˉz0z0ˉ+2i\frac{4-z_0 - \bar{z_0}}{z_0- \bar{z_0} +2i}

  1. a) π/4
  2. b) -π/2
  3. c) 3π/4
  4. d) π/2

Solution:

  1. |z - 2 + i| ≥ √5

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

    JEE Advanced Question Paper 2019 Maths Paper 1

    nk

    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:

    1. {(x, y) : xy ≤ 8, 1 ≤ y ≤ x2}

      8/x = x2

      12(x21)dx+28(8x1)dx\int_1^2 (x^2 - 1)dx + \int_2^8(\frac{8}{x} -1)dx

      = 16 loge 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?

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

    Solution:

    1. JEE Advanced Question Paper 2019 Paper 1 Maths

      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:

    E1=x29+y24=1E_1 = \frac{x^2}{9}+\frac{y^2}{4} = 1

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

    En: Ellipse x2an2+y2bn2=1\frac{x^2}{a_n^2}+\frac{y^2}{b_n^2} = 1 of largest area inscribed in Rn-1, n > 1; Rn: Rectangle of largest area, with sides parallel to the axes, inscribed in En, n > 1.

    Then which of the following options is/are correct?

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

    Solution:

    1. JEE Advanced Maths Question Paper 2019 Paper 1

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

      θ = π/4

      E

      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=12,an=3(2)n1,bn=2(2)n1r = \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 aqe

      Ea=2bn2an=3(28)×53=516E_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=[01a1233b1]M = \begin{bmatrix} 0 &1 & a \\ 1& 2 &3 \\ 3 &b &1 \end{bmatrix} and adj M = [111862531]\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?

    1. a) a + b = 3
    2. b) det(adj M2) = 81
    3. c) (adj M)-1 + adj M-1 = -M
    4. d) If M[αβγ]=[123]M \begin{bmatrix} \alpha\\ \beta\\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2\\ 3 \end{bmatrix}, then α - β + γ = 3.

    Solution:

    1. M=[01a1233b1]M = \begin{bmatrix} 0 &1 & a \\ 1& 2 &3 \\ 3 &b &1 \end{bmatrix} and adj M = [111862531]\begin{bmatrix} -1 &1 & -1 \\ 8& -6 &2 \\ -5 &3 &-1 \end{bmatrix}

      (a)

      adj M = [23bab11862b631]\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=[012123311]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[αβγ]=[123]M\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}

      This implies,

      [012123311][αβγ]=[123]\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?

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

    Solution:

    1. 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)={2x1;0x<12x28x+7;1x<3f'(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

      Math JEE Advanced Question Paper 2019 Paper 1


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

    an=αnβnαβ,n1a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, n \geq 1

    b1 = 1 and bn = an-1 + an+1, n ≥ 2.

    Then which of the following options is/are correct?

    1. a) a1+a2+a3+....+an=an+21a_1 + a_2 + a_3 +....+a_n = a_{n+2} - 1 for all n ≥ 1
    2. b) n=1an10n=1089\sum_{n=1}^\infty \frac{a_n}{10^n} = \frac{10}{89}
    3. c) n=1bn10n=889\sum_{n=1}^\infty \frac{b_n}{10^n} = \frac{8}{89}
    4. d) bn=αn+βnb_n = \alpha^n + \beta^n for all n ≥ 1.

    Solution:

    1. x2 – x – 1 = 0

      α = [1+ √5]/2 and β = [1- √5]/2

      an=αnβnαβa_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}

      As, b1 = 1 and bn = an-1 + an+1, n ≥ 2

      Solved JEE Advanced Question Paper 2019 Paper 1 Maths

      a1 + a2 + …+ an = n=1an10n\sum_{n=1}^\infty \frac{a_n}{10^n}

      => an + an+1 = an+2

      => (α10)n(β10)nαβ\frac{\sum (\frac{\alpha}{10})^n - \sum(\frac{\beta}{10})^n}{\alpha - \beta}

      r=1nar=an+2a2\sum_{r=1}^n a_r= a_{n+2} - a_2

      = an+2 – [α2 - β2]/[α -β]

      = an+2 – [α + β]

      = an+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 YP. If PYP has length 1 for each point P on , then which of the following is options is/are correct?

    1. a) y=loge(1+1x2x)1x2y = log_e(\frac{1+\sqrt{1-x^2}}{x}) - \sqrt{1-x^2}
    2. b) xy' - √[1-x2] = 0
    3. c) y=loge(1+1x2x)+1x2y = log_e(\frac{1+\sqrt{1-x^2}}{x}) + \sqrt{1-x^2}
    4. d) xy' + √[1-x2] = 0

    Solution:

    1. Solved JEE Advanced Maths Question Paper 2019 Paper 1

      y - y1 = m(x – x1)

      yp - 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=±1x2x dxy = \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=loge(1+1x2x)1x2y = 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?

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

    Solution:

    1. (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

      Answers for JEE Advanced Maths Question Paper 1


    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:

      1. L1:r=i^+λ(i^+2j^+2k^)L_1 : \vec r = \hat i + \lambda(- \hat i + 2 \hat j + 2 \hat k)

        L2:r=μ(2i^j^+2k^)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, 6i^+6j^3k^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)

        =>2μ+λ12=μ2λ2=2μ2λ1=k\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.

        So, possible vectors of A are (8/9 i^,2/9 j^,2/9 k^) OR (4/9 i^,2/9 j^,4/9 k^) OR (2/3 i^+1/3 k^)(8/9 \ \hat i, 2/9 \ \hat j, 2/9\ \hat k) \ OR \ (4/9 \ \hat i, -2/9 \ \hat j, 4/9 \ \hat k) \ OR \ (2/3 \ \hat i + 1/3 \ \hat k)

        Answer: a, b, d


      Question 13: If I=2ππ/4π/4dx(1+esin x)(2cos2x)I = \frac{2}{\pi} \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{sin\ x})(2-cos 2x)} then 27I2 equals _________

        Solution:

        1. Solved JEE Advanced Maths 2019 Question Paper 1

        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:

          1. Solved JEE Advanced Math 2019 Question Paper 1

            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_________

          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

            Solution:

            1. 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:

              1. S: 29

                P(E1E2)=P(E1E2)P(E2)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

                111111100\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(E1E2)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. 1st line: x = λ, y = 0, z = 0

                  x + y + z = 1 => λ = 1

                  A(1, 0, 0)

                  2nd 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 1 be a cube root of unity. Then the minimum of the set {|a + b + c2|2 ; Where, a, b, c distinct non-zero integers }equals ________

                  Solution:

                  1. | a + b + c2|2 = (a + b + c2) (a + c + b2)

                    = (a2 + b2 + c2 - ab – bc – ca)

                    [because, conjugate of (a + b + c2) = (a + c + b2)]

                    = > (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]


                  Video Lessons - Paper 1 Maths

                  JEE Advanced 2019 Maths Paper 1 Solutions

                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions
                  JEE Advanced 2019 Maths Paper 1 Question and Solutions

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