JEE Advanced 2020 Maths Paper 1

Here students will get access to the complete set of questions from the Paper 1 Maths section of the JEE Advanced 2020 question paper. 18 questions were asked and a total of 66 marks were allocated for this section. There were MCQs, Numerical types question as well as paragraph based questions. The question paper provided here will help students get a better understanding of the question pattern, marking scheme and duration of the paper. Students can either view the paper directly or download it in a PDF format. They can use the question paper as a study guide and practice solving different problems. Detailed solutions are also provided to help the students find the right answer and study effectively for the entrance exam.

Paper 1 - Maths

Question 1. Suppose 𝑎, 𝑏 denote the distinct real roots of the quadratic polynomial 𝑥2 + 20𝑥 − 2020 and suppose 𝑐, 𝑑 denote the distinct complex roots of the quadratic polynomial

x2 − 20𝑥 + 2020. Then the value of 𝑎𝑐 (𝑎 − 𝑐) + 𝑎𝑑 (𝑎 − 𝑑) + 𝑏𝑐 (𝑏 − 𝑐) + 𝑏𝑑 (𝑏 − 𝑑) is

  1. a) 0
  2. b) 8000
  3. c) 8080
  4. d) 16000

Solution:

  1. Answer: d

    JEE Advanced Paper 1 Maths Question 1 Solution

    Now

    = ac (a – c) + ad (a – d) +bc (b – c) + bd (b – d)

    = a2c - ac2 + a2d - ad2 +b2c – bc2+ b2d - bd2 

    = a2 (c +d) + b2 (c + d) – c2 (a + b) – d2 (a + b)

    = (a2 + b2) (c + d) – (a + b) (c2 + d2)

    = {(a+b)2 – 2ab} (c + d) – {(c + d)2 –2cd} (a + b)

    Put value a, b, c and d then,

    = {(-20)2 – 2(-2020)} (20) - {(20)2 – 2(2020)} (-20)}

    = ((400 + 4040) (20) – (–20) ((20)2– 4040)

    = 20[4440 – 3640]

    20[800] = 16000


Question 2. If the function 𝑓: R ⟶ R is defined by (𝑥) = |𝑥| (𝑥– sin𝑥), then which of the following statements is TRUE?

  1. a) 𝑓 is one-one, but NOT onto
  2. b) 𝑓 is onto, but NOT one-one
  3. c) 𝑓 is BOTH one-one and onto
  4. d) 𝑓 is NEITHER one-one NOR onto

Solution:

  1. Answer: c

    Given, f(x) = |x|(x – sinx)

    f(–x) = –(|x|(x – sinx))

    f(–x) = –f(x) ⇒ f(x) is odd, non-periodic and continuous function.

    Now,

    JEE Advanced Paper 1 Maths Question 2 Solution


Question 3. Let the functions: R ⟶ R and g : R ⟶ R be defined by f(x)=ex1ex1andg(x)=12(ex1+e1x)f(x) = e^{x-1} - e^{-\left | x-1 \right |}\: and\: g(x) = \frac{1}{2}(e^{x-1} + e^{1-x}) Then, the area of the region in the first quadrant bounded by the curves 𝑦 = (𝑥), 𝑦 = g(𝑥) and 𝑥 = 0 is

  1. a) (23)+12(ee1)(2-\sqrt{3})+\frac{1}{2}(e-e^{-1})
  2. b) (2+3)+12(ee1)(2+\sqrt{3})+\frac{1}{2}(e-e^{-1})
  3. c) (23)+12(e+e1)(2-\sqrt{3})+\frac{1}{2}(e+e^{-1})
  4. d) (2+3)+12(e+e1)(2+\sqrt{3})+\frac{1}{2}(e+e^{-1})

Solution:

  1. Answer: a

    JEE Advanced Paper 1 Maths Question 3 Solution

    JEE Advanced Paper 1 Maths Question 3 Solution


Question 4. Let 𝑎, 𝑏 and λ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola 𝑦2 = 4λ𝑥, and suppose the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1 passes through the point𝑃. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is

  1. a) 1/√2
  2. b) 1/2
  3. c) 1/3
  4. d) 2/5

Solution:

  1. Answer: a


    JEE Advanced Paper 1 Maths Question 4 Solution

    P(λ, 2λ)

    y2=4λx(dydx)A=1=m1y^{2} = 4\lambda x \Rightarrow \left ( \frac{dy}{dx} \right )_{A} = 1 = m_{1}…….(1)

    Now E: x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1 Passes through P

    λ2a2+4λ2b2=1(dydx)A=b22a2=m2\frac{\lambda ^{2}}{a^{2}}+\frac{4\lambda ^{2}}{b^{2}} = 1 \Rightarrow \left ( \frac{dy}{dx} \right )_{A} = \frac{-b^{2}}{2a^{2}} = m_{2}

    2λ2λXλa2b22λ=1\frac{2\lambda }{2\lambda }X - \frac{\lambda }{a^{2}}\cdot \frac{b^{2}}{2\lambda } = -1 …….(2)

    From eq. (1) and (2)

    m1. m2 = -1 ⇒ b2 = 2a2

    a2b2=12\frac{a^{2}}{b^{2}}=\frac{1}{2}

    for eccentricity of ellipse,

    e=1a2b2=112=12e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}


Question 5. Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are23and13\frac{2}{3} and \frac{1}{3}, respectively. Suppose α is the number of heads that appear when C1 is tossed twice, independently, and suppose α is the number of heads that appear when C2 is tossed twice, independently. Then the probability that the roots of the quadratic polynomial 𝑥2 − 𝛼𝑥 + 𝛽 are real and equal, is

  1. a) 40/81
  2. b) 20/81
  3. c) 1/2
  4. d) 1/4

Solution:

  1. Answer: b

    JEE Advanced Paper 1 Maths Question 5 Solution

    Now roots of equation x2– αx + β = 0 are real and equal

    ∴ D= 0

    α2 – 4β = 0

    α2 = 4β

    ⇒ (α = 0, β = 0) or (α = 2, β = 1)

    P(E)=2Co(13)22Co(23)2+2C2(23)22C1(13)(23)P(E) = ^{2}C_{o}\left ( \frac{1}{3} \right )^{2}\cdot ^{2}C_{o}\left ( \frac{2}{3} \right )^{2} + ^{2}C_{2}\left ( \frac{2}{3} \right )^{2}\cdot ^{2}C_{1}\left ( \frac{1}{3} \right )\left ( \frac{2}{3} \right )

    19×49+49×49=2081\frac{1}{9}\times\frac{4}{9}+\frac{4}{9}\times\frac{4}{9}=\frac{20}{81}


Question 6. Consider all rectangles lying in the region {(x,y)ϵR×R:0xπ2and0y2sin(2x)}\left \{ (x,y)\epsilon R\times R:0 \leq x\leq \frac{\pi }{2}\: and\: 0\leq y\leq 2sin(2x)\right \} and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

  1. a) 3π2\frac{3\pi }{2}
  2. b) 𝜋
  3. c) π23\frac{\pi }{2\sqrt{3}}
  4. d) π32\frac{\pi\sqrt{3} }{2}

Solution:

  1. Answer: c

    JEE Advanced Paper 1 Maths Question 6 Solution

    Let sides of rectangle are a & b

    Then, perimeter = 2a + 2b

    p = 2(a + b)

    Now, b = 2sin2x and b = 2sin(2x + 2a) ⇒ 2x + 2x + 2a = 𝜋

    {x=π4a2}\left \{ x = \frac{\pi }{4}-\frac{a}{2}\right \}

    For perimeter maximum

    P = 2a + 2b

    P = 𝜋 – 4x + 4sin2x

    dpdx=4+8cos2x=8{cos2x12}\frac{dp}{dx} = -4+8cos2x = 8\left \{cos2x-\frac{1}{2} \right \}

    JEE Advanced Paper 1 Maths Question 7 Solution


Question 7. Let the function 𝑓: R → R be defined by (𝑥) = 𝑥3 − 𝑥2 + (𝑥 − 1) sin 𝑥 and let 𝑔: R → R be an arbitrary function. Let 𝑓g : R → R be the product function defined by (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥). Then which of the following statements is/are TRUE?

  1. a) If g is continuous at 𝑥 = 1, then 𝑓𝑔 is differentiable at 𝑥 = 1
  2. b) If 𝑓𝑔 is differentiable at 𝑥 = 1, then g is continuous at 𝑥 = 1
  3. c) If g is differentiable at 𝑥 = 1, then 𝑓𝑔 is differentiable at 𝑥 = 1
  4. d) If 𝑓𝑔 is differentiable at 𝑥 = 1, then g is differentiable at 𝑥 = 1

Solution:

  1. Answer: a, c

    f : R → R

    a) f(x) = x3– x2+ (x – 1) sinx ; g : R → R

    h(x) = f(x). g(x) = {x3 – x2+ (x – 1)sinx}. g(x)

    JEE Advanced Paper 1 Maths Question 7 Solution

    as g(x) is constant at x =1

    ∴ g(1+h) = g(1 – h) = g(1)

    h'(1+) = h'(1-) = (1 + sin1) g(1)

    'a' is correct.


Question 8. Let M be a 3 × 3 invertible matrix with real entries and let I denote the 3 × 3 identity matrix. If 𝑀−1 = adj (adj 𝑀), then which of the following statements is/are ALWAYS TRUE?

  1. a) 𝑀 = 𝐼
  2. b) det 𝑀 = 1
  3. c) 𝑀2 = 𝐼
  4. d) (adj 𝑀)2 = 𝐼

Solution:

  1. Answer: b, c, d

    M–1 = adj(adj(M))

    (adj M)M–1 = (adjM)(adj(adj(M)))

    (adj M)M–1 = N. adj(N) { Let adj(M) = N }

    (adj M)M–1 = |N|I

    (adjM)M–1 = |adj(M)|I3

    (adjM) = |M|2 .M ............(1)

    |adj M| = ||M|2.M|

    |M|2 = |M6|.|M|

    |M|=1, |M| ≠ 0

    From equation (1)

    adj.M =M ............(2)

    Multiply by matrix M

    M.adj M = M2

    |M|I3 = M2

    M2 = I

    From (2) adj M =M

    (adj M)2 = M2 = I


Question 9. Let S be the set of all complex numbers z satisfying |𝑧2 + 𝑧 + 1| = 1. Then which of the following statements is/are TRUE?

JEE Advanced Paper 1 Maths Question 9 Options

    Solution:

    1. Answer: b, c

      JEE Advanced Paper 1 Maths Question 9 Solution


    Question 10. Let x, 𝑦 and z be positive real numbers. Suppose 𝑥, 𝑦 and z are the lengths of the sides of a triangle opposite to its angles 𝑋, 𝑌 and Z, respectively. If tanX2+tanZ2=2yx+y+ztan\frac{X}{2}+tan\frac{Z}{2} =\frac{2y}{x+y+z}, then which of the following statements is/are TRUE?

    1. a) 2𝑌 = 𝑋 + 𝑍
    2. b) Y = 𝑋 + 𝑍
    3. c) tanX2=xy+ztan\frac{X}{2} =\frac{x}{y+z}
    4. d) 𝑥2 + 𝑧2 − 𝑦2 = 𝑥𝑧

    Solution:

    1. Answer: b, c

      JEE Advanced Paper 1 Maths Question 10 Solution

      2 = (S - x)2(S - z)2

      S(S–y) = (S–x) (S–z)

      (x + y + z) (x + z – y) = (y + z – x) (x + y –z)

      (x + z)2 – y2 = y2 - (z – x)2

      (x + z)2 + (x – z)2 = 2y2

      x 2 + z2 = y2

      JEE Advanced Paper 1 Maths Question 10 Solution


    Question 11. Let 𝐿1 and 𝐿2 be the following straight lines.

    L1:x11=y1=z13L_{1}:\frac{x-1}{1} = \frac{y}{1} = \frac{z-1}{3} and L2:x13=y1=z11L_{2}:\frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}

    Suppose the straight line is

    L:xαl=y1m=zγ2L:\frac{x-\alpha }{l} = \frac{y-1}{m} = \frac{z-\gamma }{-2}

    Lies in the plane containing 𝐿1 and 𝐿2, and passes through the point of intersection of 𝐿1 and

    𝐿2. If the line L bisects the acute angle between the lines 𝐿1 and 𝐿2, then which of the following statements is/are TRUE?

    1. a) 𝛼 − 𝛾 = 3
    2. b) l + 𝑚 = 2
    3. c) 𝛼 − 𝛾 = 1
    4. d) 𝑙 + 𝑚 = 0

    Solution:

    1. Answer: a, b

      Solution:JEE Advanced Paper 1 Maths Question 11 Solution

      JEE Advanced Paper 1 Maths Question 11 Solution


    Question 12. Which of the following inequalities is/are TRUE?

    1. a)01xcosxdx38\int_{0}^{1}xcosx\: dx\geq \frac{3}{8}
    2. b) 01xsinxdx310\int_{0}^{1}xsinx\: dx\geq \frac{3}{10}
    3. c) 01x2cosxdx12\int_{0}^{1}x^{2}cosx\: dx\geq \frac{1}{2}
    4. d) 01x2sinxdx29\int_{0}^{1}x^{2}sinx\: dx\geq \frac{2}{9}

    Solution:

    1. Answer: a, b, d

      JEE Advanced Paper 1 Maths Question 12 Solution

      JEE Advanced Paper 1 Maths Question 12 Solution

      JEE Advanced Paper 1 Maths Question 12 Solution


    Question 13. Let m be the minimum possible value of log3(3y1+3y2+3y3)log_{3}(3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}), where 𝑦1, 𝑦2, 𝑦3 are real numbers for which 𝑦1 + 𝑦2 + 𝑦3 = 9. Let M be the maximum possible value of (log3𝑥1 + log3𝑥2 + log3𝑥3), where x1, 𝑥2, 𝑥3 are positive real numbers for which 𝑥1 + 𝑥2 + 𝑥3 = 9. Then the value of log2(𝑚3) + log3(𝑀2) is _____

      Solution:

      1. Answer: 8.00

        Using AM ≥ GM

        3y1+3y2+3y33(3y13y23y3)13\frac{3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}{3}\geq (3^{y_{1}}\cdot 3^{y_{2}}\cdot 3^{y_{3}})^{\frac{1}{3}}

        3y1+3y2+3y33(3y1+y2+y3)13{3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}\geq 3\cdot (3^{y_{1}+y_{2}+y_{3}})^{\frac{1}{3}} {∴ y1+ y2 + y3=9}

        3y1+3y2+3y33(39)13{3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}\geq 3\cdot (3^{9})^{\frac{1}{3}}

        3y1+3y2+3y381{3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}\geq 81

        m = log381 = log334 = 4log33 = 4

        Again, using A.M ≥ G.M

        x1+x2+x33(x1x2x3)13\frac{x_{1}+x_{2}+x_{3}}{3}\geq (x_{1}\cdot x_{2}\cdot x_{3})^{\frac{1}{3}}

        = 93(x1x2x3)13\frac{9}{3}\geq (x_{1}\cdot x_{2}\cdot x_{3})^{\frac{1}{3}} { ∵ x1+x2+x3 = 9}

        ⇒27 ≥ x1x2x3

        M = log3𝑥1+ log3𝑥2+ log3𝑥3

        M = log3(x1x2x3) = log3(27) =3

        ∴ log2(m)3 + log3(M)2 ⇒ log2(26) + log3(32) = 6 + 2 = 8


      Question 14. Let 𝑎1, 𝑎2, 𝑎3,… be a sequence of positive integers in arithmetic progression with common difference 2. Also, let 𝑏1, b2, 𝑏3,… be a sequence of positive integers in geometric progression with common ratio 2. If a1 = 𝑏1 = 𝑐, then the number of all possible values of c, for which the equality 2(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛) = 𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 holds for some positive integer n, is _____

        Solution:

        1. Answer: 1.00

          2(a1+a2 +......+ an) = b1 + b2 + ... + bn

          2[n2(2a1+(n1)2)]=b1(2n1)212\left [ \frac{n}{2}(2a_{1} +(n-1)2) \right ] = \frac{b_{1}(2^{n}-1)}{2-1}

          ⇒ 2n[a1 + (n – 1)] = b1(2n – 1)

          ⇒ 2na1+ 2n2 – 2n = a1(2n – 1) {∵ a1 = b1}

          ⇒ 2n2 – 2n = a1(2 – 1=2n)

          a1=2(n2n)2n12n=ca_{1} = \frac{2(n^{2}-n)}{2^{n}-1-2n} = c {∵ a1 = c}

          ∵ c ≥ 1

          2(n2n)2n12n1\Rightarrow \frac{2(n^{2}-n)}{2^{n}-1-2n}\geq 1

          2(n2– n) ≥ 2n– 1 – 2n {∵ n2– n ≥ 0 for n ≥ 1}

          = 2n2+ 1 ≥ 2n

          Therefore, n =1, 2,3,4,5,6

          n = 1 ⇒ c= 0 (rejected)

          n = 2 ⇒ c < 0 (rejected)

          n = 3 ⇒ c = 12 (correct)

          n = 4 ⇒ c = not Integer

          n = 5 ⇒ c = not Integer

          n = 6 ⇒ c = not Integer

          ∵ c = 12 for n = 3

          Hence, no. of such c = 1


        Question 15. Let f: [0, 2] ⟶ R be the function defined by f(x)=(3sin(2πx))sin(πxπ4)sin(3πx+π4)f(x) = (3-sin(2\pi x))sin\left (\pi x-\frac{\pi }{4} \right) - sin\left ( 3\pi x +\frac{\pi }{4} \right )

        If 𝛼, 𝛽 ∈ [0,2] are such that {𝑥 ∈ [0, 2] ∶ f(𝑥) ≥ 0} = [𝛼, 𝛽], then the value of 𝛽 − 𝛼 is _____

          Solution:

          1. Answer: 1.00

            f(x)=(3sin(2πx))sin(πxπ4)sin(3πx+π4)f(x) = (3-sin(2\pi x))sin\left (\pi x-\frac{\pi }{4} \right) - sin\left ( 3\pi x +\frac{\pi }{4} \right )

            JEE Advanced Paper 1 Maths Question 15 Solution


          Question 16. In a triangle PQR, let a=QR,b=RPandc=PQ\vec{a} = \vec{QR}, \vec{b} = \vec{RP}\: and \: \vec{c} = \vec{PQ}. If

          a=3|\vec{a}| = 3, and b=4anda.(cb)c.(ab)=aa+b|\vec{b}| = 4\: and\: \frac{\vec{a}.(\vec{c} - \vec{b})}{\vec{c}.(\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}|+|\vec{b}|}

          Then the value of is _____

            Solution:

            1. Answer: 108.00

              JEE Advanced Paper 1 Maths Question 16 Solution


            Question 17. For a polynomial g(𝑥) with real coefficients, let 𝑚𝑔 denote the number of distinct real roots of g(𝑥). Suppose S is the set of polynomials with real coefficients defined by S = {(x2 – 1)2(a0 + a1x + a2x2 + a3x3) : a0, a1, a2, a3∈R}

            For a polynomial 𝑓, let 𝑓'and 𝑓'' denote its first and second order derivatives, respectively.

            Then the minimum possible value of (𝑚𝑓' + 𝑚𝑓''), where 𝑓 ∈ 𝑆, is _____

              Solution:

              1. Answer: 5.00

                f(x) = (x2 – 1)2h(x); h(x) = a0 +a1x +a2x2+ a3x3

                Now, f(1) = f(–1) = 0

                ⇒ f'(α) = 0, α∈ (–1,1) [Rolle's Theorem]

                Also, f'(1) = f'(–1) = 0 ⇒ f'(x) = 0 has at least 3 root –1, α,1 with –1 <α< 1

                ⇒ f"(x) = 0 will have at least 2 roots, say β, γ such that

                –1 < β < α < γ < 1 [Rolle's Theorem]

                So, min= 2

                and we find (mf' + mf") = 5 for f(x) =(x2 – 1)2

                Thus, Ans = 5


              Question 18. Let e denote the base of the natural logarithm. The value of the real number a for which the right-hand limit is equal to a nonzero real number, is _____

                Solution:

                1. Answer: 1.00

                  JEE Advanced Paper 1 Maths Question 18 Solution


                Video Lessons - Paper 1 Maths

                JEE Advanced 2020 Maths Paper 1 Solutions

                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions
                JEE Advanced 2020 Maths Paper 1 Question and Solutions