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 π‘Ž, 𝑏 denote the distinct real roots of the quadratic polynomial π‘₯2 + 20π‘₯ βˆ’ 2020 and suppose 𝑐, 𝑑 denote the distinct complex roots of the quadratic polynomial

x2 βˆ’ 20π‘₯ + 2020. Then the value of π‘Žπ‘ (π‘Ž βˆ’ 𝑐) + π‘Žπ‘‘ (π‘Ž βˆ’ 𝑑) + 𝑏𝑐 (𝑏 βˆ’ 𝑐) + 𝑏𝑑 (𝑏 βˆ’ 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)

    =Β a2c - ac2 + a2dΒ - ad2Β +b2c – bc2+ b2d - bd2Β 

    = 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 (π‘₯) = |π‘₯| (π‘₯– sinπ‘₯), 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) β‡’ 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

\(f(x) = e^{x-1} - e^{-\left | x-1 \right |}\: and\: g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})\)
Then, the area of the region in the first quadrant bounded by the curves 𝑦 = (π‘₯), 𝑦 = g(π‘₯) and π‘₯ = 0 is

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

Solution:

  1. Answer: a

    JEE Advanced Paper 1 Maths Question 3 Solution

    JEE Advanced Paper 1 Maths Question 3 Solution


Question 4. Let π‘Ž, 𝑏 and Ξ» be positive real numbers. Suppose P is an end point of the latus rectum of the parabola 𝑦2 = 4Ξ»π‘₯, and suppose the ellipse

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\)
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(Ξ», 2Ξ»)

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

    Now E:

    \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\)
    Passes through P

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

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

    From eq. (1) and (2)

    m1. m2 = -1 β‡’ b2 = 2a2

    \(\frac{a^{2}}{b^{2}}=\frac{1}{2}\)

    for eccentricity of ellipse,

    \(e = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}\)


Question 5. Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are

\(\frac{2}{3} and \frac{1}{3}\)
, respectively. SupposeΒ Ξ± is the number of heads that appear when C1 is tossed twice, independently, and supposeΒ Ξ± is the number of heads that appear when C2 is tossed twice, independently. Then the probability that the roots of the quadratic polynomial π‘₯2 βˆ’ 𝛼π‘₯ + 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– Ξ±x +Β Ξ² = 0 are real and equal

    ∴ D= 0

    Ξ±2 – 4Ξ² = 0

    Ξ±2 = 4Ξ²

    β‡’ (Ξ± = 0,Β Ξ² = 0) or (Ξ± = 2,Β Ξ² = 1)

    \(P(E) = ^{2}C_{o}\left ( \frac{1}{3} \right )^{2}\cdot ^{2}C_{o}\left ( \frac{2}{3} \right )^{2} + ^{2}C_{2}\left ( \frac{2}{3} \right )^{2}\cdot ^{2}C_{1}\left ( \frac{1}{3} \right )\left ( \frac{2}{3} \right )\)

    β‡’

    \(\frac{1}{9}\times\frac{4}{9}+\frac{4}{9}\times\frac{4}{9}=\frac{20}{81}\)


Question 6. Consider all rectangles lying in the region

\(\left \{ (x,y)\epsilon R\times R:0 \leq x\leq \frac{\pi }{2}\: and\: 0\leq y\leq 2sin(2x)\right \}\)
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)
    \(\frac{3\pi }{2}\)
  2. b) πœ‹
  3. c)
    \(\frac{\pi }{2\sqrt{3}}\)
  4. d)
    \(\frac{\pi\sqrt{3} }{2}\)

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) β‡’ 2x + 2x + 2a =Β πœ‹

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

    For perimeter maximum

    P = 2a + 2b

    P = πœ‹ – 4x + 4sin2x

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

    JEE Advanced Paper 1 Maths Question 7 Solution


Question 7. Let the function 𝑓: R β†’ R be defined by (π‘₯) = π‘₯3 βˆ’ π‘₯2 + (π‘₯ βˆ’ 1) sin π‘₯ and let 𝑔: R β†’ R be an arbitrary function. Let 𝑓g : R β†’ R be the product function defined by (𝑓𝑔)(π‘₯) = 𝑓(π‘₯)𝑔(π‘₯). Then which of the following statements is/are TRUE?

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

Solution:

  1. Answer: a, c

    f : R β†’ R

    a) f(x) = x3– x2+ (x – 1) sinx ; g : R β†’ 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 Γ— 3 invertible matrix with real entries and let I denote the 3 Γ— 3 identity matrix. If π‘€βˆ’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| β‰  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 π‘₯, 𝑦 and z are the lengths of the sides of a triangle opposite to its angles 𝑋, π‘Œ and Z, respectively. If

    \(tan\frac{X}{2}+tan\frac{Z}{2} =\frac{2y}{x+y+z}\)
    , then which of the following statements is/are TRUE?

    1. a) 2π‘Œ = 𝑋 + 𝑍
    2. b) Y = 𝑋 + 𝑍
    3. c)
      \(tan\frac{X}{2} =\frac{x}{y+z}\)
    4. d) π‘₯2 + 𝑧2 βˆ’ 𝑦2 = π‘₯𝑧

    Solution:

    1. Answer: b, c

      JEE Advanced Paper 1 Maths Question 10 Solution

      βˆ†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.

    \(L_{1}:\frac{x-1}{1} = \frac{y}{1} = \frac{z-1}{3}\)
    and
    \(L_{2}:\frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}\)

    Suppose the straight line is

    \(L:\frac{x-\alpha }{l} = \frac{y-1}{m} = \frac{z-\gamma }{-2}\)

    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 βˆ’ 𝛾 = 3
    2. b) l + π‘š = 2
    3. c) 𝛼 βˆ’ 𝛾 = 1
    4. d) 𝑙 + π‘š = 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)
      \(\int_{0}^{1}xcosx\: dx\geq \frac{3}{8}\)
    2. b)
      \(\int_{0}^{1}xsinx\: dx\geq \frac{3}{10}\)
    3. c)
      \(\int_{0}^{1}x^{2}cosx\: dx\geq \frac{1}{2}\)
    4. d)
      \(\int_{0}^{1}x^{2}sinx\: dx\geq \frac{2}{9}\)

    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

    \(log_{3}(3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}})\)
    , where 𝑦1, 𝑦2, 𝑦3 are real numbers for which 𝑦1 + 𝑦2 + 𝑦3 = 9. Let M be the maximum possible value of (log3π‘₯1 + log3π‘₯2 + log3π‘₯3), where x1, π‘₯2, π‘₯3 are positive real numbers for which π‘₯1 + π‘₯2 + π‘₯3 = 9. Then the value of log2(π‘š3) + log3(𝑀2) is _____

      Solution:

      1. Answer: 8.00

        Using AM β‰₯ GM

        \(\frac{3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}{3}\geq (3^{y_{1}}\cdot 3^{y_{2}}\cdot 3^{y_{3}})^{\frac{1}{3}}\)

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

        \({3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}\geq 3\cdot (3^{9})^{\frac{1}{3}}\)

        \({3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}}\geq 81\)

        m = log381 = log334 = 4log33 = 4

        Again, using A.M β‰₯ G.M

        \(\frac{x_{1}+x_{2}+x_{3}}{3}\geq (x_{1}\cdot x_{2}\cdot x_{3})^{\frac{1}{3}}\)

        =

        \(\frac{9}{3}\geq (x_{1}\cdot x_{2}\cdot x_{3})^{\frac{1}{3}}\)
        { ∡ x1+x2+x3 = 9}

        β‡’27Β β‰₯ x1x2x3

        M = log3π‘₯1+ log3π‘₯2+ log3π‘₯3

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

        ∴ log2(m)3 + log3(M)2 β‡’ log2(26) + log3(32) = 6 + 2 = 8


      Question 14. Let π‘Ž1, π‘Ž2, π‘Ž3,… 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(π‘Ž1 + π‘Ž2 + β‹― + π‘Žπ‘›) = 𝑏1 + 𝑏2 + β‹― + 𝑏𝑛 holds for some positive integer n, is _____

        Solution:

        1. Answer: 1.00

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

          \(2\left [ \frac{n}{2}(2a_{1} +(n-1)2) \right ] = \frac{b_{1}(2^{n}-1)}{2-1}\)

          β‡’ 2n[a1 + (n – 1)] = b1(2n – 1)

          β‡’ 2na1+ 2n2 – 2n = a1(2n – 1) {∡ a1 = b1}

          β‡’ 2n2 – 2n = a1(2 – 1=2n)

          \(a_{1} = \frac{2(n^{2}-n)}{2^{n}-1-2n} = c\)
          {∡ a1 = c}

          ∡ cΒ β‰₯ 1

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

          2(n2– n)Β β‰₯ 2n– 1 – 2n {∡ n2– nΒ β‰₯ 0 for nΒ β‰₯ 1}

          = 2n2+ 1Β β‰₯ 2n

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

          n = 1 β‡’ c= 0 (rejected)

          n = 2 β‡’ c < 0 (rejected)

          n = 3 β‡’ c = 12 (correct)

          n = 4Β β‡’ c = not Integer

          n = 5 β‡’ c = not Integer

          n = 6Β β‡’ 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

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

        If 𝛼, 𝛽 ∈ [0,2] are such that {π‘₯ ∈ [0, 2] ∢ f(π‘₯) β‰₯ 0} = [𝛼, 𝛽], then the value of 𝛽 βˆ’ 𝛼 is _____

          Solution:

          1. Answer: 1.00

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

            JEE Advanced Paper 1 Maths Question 15 Solution


          Question 16. In a triangle PQR, letΒ 

          \(\vec{a} = \vec{QR}, \vec{b} = \vec{RP}\: and \: \vec{c} = \vec{PQ}\)
          . If

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

          Then the value of is _____

            Solution:

            1. Answer: 108.00

              JEE Advanced Paper 1 Maths Question 16 Solution


            Question 17. For a polynomial g(π‘₯) with real coefficients, let π‘šπ‘” denote the number of distinct real roots of g(π‘₯). 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 (π‘šπ‘“'Β + π‘šπ‘“''), 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

                β‡’ f'(Ξ±) = 0, α∈ (–1,1) [Rolle's Theorem]

                Also, f'(1) = f'(–1) = 0 β‡’ f'(x) = 0 has at least 3 root –1, Ξ±,1 with –1 <Ξ±< 1

                β‡’ f"(x) = 0 will have at least 2 roots, sayΒ Ξ², Ξ³ such that

                –1 < Ξ² < Ξ± < Ξ³ < 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