JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1

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JEE Main 2021 24th February Shift 1 Maths Question Paper

SECTION A


Question 1: The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is:

  1. a. x = a
  2. b. x = 0
  3. c. x = -a/2
  4. d. x = a/2

Solution:

  1. Answer: (b)

    JEE Main 2021 Paper With Solution Maths Feb 24 Shift 1

    h = (at2+a)/2, k = (2at+0)/2

    ⇒ t2 =(2h-a)/a and t=k/a

    ⇒ k2/a2 =(2h-a)/a

    ⇒ Locus of (h, k) is y2 = a (2x – a)

    ⇒ y2 = 2a(x- a/2)

    Its directrix is x - a/2 = -a/2

    ⇒ x = 0


Question 2: A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

  1. a. 560
  2. b. 1050
  3. c. 1625
  4. d. 575

Solution:

  1. Answer: (c)

    (2I,4F)+ (3I,6F) + (4I,8F)

    = 6C2 8C4+6C3 8C6+6C4 8C8^6C_2\ ^8C_4 + ^6C_3\ ^8C_6 + ^6C_4\ ^8 C_8

    = 15 × 70 + 20 × 28 + 15 × 1

    = 1050 + 560 + 15 = 1625


Question 3: The equation of the plane passing through the point (1,2,–3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is:

  1. a. 3x – 10y – 2z + 11 = 0
  2. b. 6x – 5y – 2z – 2 = 0
  3. c. 11x + y + 17z + 38 = 0
  4. d. 6x – 5y + 2z + 10 = 0

Solution:

  1. Answer: (c)

    JEE Main 2021 Papers With Solutions Feb 24 Maths Shift 1


Question 4: A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 1/4. Three stones A, B and C are placed at the points (1,1),(2,2), and (4,4) respectively. Then which of these stones is/are on the path of the man?

  1. a. B only
  2. b. A only
  3. c. the three
  4. d. C only

Solution:

  1. Answer: (a)

    x/a + y/b = 1

    h/a + k/b = 1 ⋯⋯(1)

    and (1/a + 1/b)/2 = 1/4

    ∴1/a + 1/b = 1/2 ⋯⋯(2)

    (From (1) and (2))

    Line passes through fixed point B(2, 2)


Question 5: The statement among the following that is a tautology is:

  1. a. A∧(A∨B)
  2. b. B→[A∧(A→B)]
  3. c. A∨(A∧B)
  4. d. [A∧(A→B)]→B

Solution:

  1. Answer: (d)

    A∧ (~ A∨B)→B

    = [(A∧~A)∨(A∧B)]→ B

    = (A∧ B)→ B

    = ~ (A∧B)∨B

    = t


Question 6: Let f :R→R be defined as f(x) = 2x-1 and g:R - {1} →R be defined as g(x) = (x-½)/(x-1).

Then the composition function f(g(x)) is:

  1. a. Both one-one and onto
  2. b. onto but not one-one
  3. c. Neither one-one nor onto
  4. d. one-one but not onto

Solution:

  1. Answer: (d)

    f(g(x)) = 2g(x) - 1

    = 2(x - 1/2)/(x - 1) - 1

    = x/(x - 1)

    f(g(x)) = 1 + 1/(x - 1)

    ∴ one-one, into


Question 7: If f:R→ R is a function defined by f(x) = [x - 1] cos ((2x - 1)/2)𝜋 , where [.] denotes the greatest integer function, then f is:

  1. a. discontinuous only at x = 1
  2. b. discontinuous at all integral values of x except at x = 1
  3. c. continuous only at x = 1
  4. d. continuous for every real x

Solution:

  1. Answer: (d)

    Doubtful points are x = n, n∈I

    JEE Main 2021 Maths Papers Feb 24 Shift 1 With Solutions

    f(n) = 0

    Hence continuous.


Question 8: The function f(x) = (4x3- 3x2)/6 - 2sin x + (2x - 1)cos x:

  1. a. increases in [1/2, ∞)
  2. b. decreases (-∞, 1/2]
  3. c. increases in (-∞, 1/2]
  4. d. decreases [1/2, ∞)

Solution:

  1. Answer: (a)

    f’(x) = (2x - 1) (x - sin x )

    ⇒ f'x ≥ 0 in x∈(-∞, 0] ⋃ [1/2, ∞)

    and f'x ≤ 0 in x∈(0, ½)


Question 9: The distance of the point (1, 1, 9) from the point of intersection of the line (x - 3)/1 = (y - 4)/2 = (z - 5)/2 and the plane x + y + z = 17 is:

  1. a. √38
  2. b. 19√2
  3. c. 2√19
  4. d. 38

Solution:

  1. Answer: (a)

    (x - 3)/1 = (y - 4)/2 = (z - 5)/2 = λ

    x = λ + 3, y = 2λ+ 4, z = 2λ+5

    Which lies on given plane hence

    ⇒ λ+3+2λ +4+2λ+5 = 17

    ⇒ λ = 5/5 = 1

    Hence, point of intersection is Q (4, 6, 7)

    ∴ Required distance =PQ

    = √(9+25+4)

    = √38


Question 10:


JEE Main Solution 2021 Maths Papers Feb 24 Shift 1

  1. a. 2/3
  2. b. 0
  3. c. 1/15
  4. d. 3/2

Solution:

  1. Answer: (a)

    JEE Main Solution Feb 24 Shift 1 2021 Maths Papers


Question 11: Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

  1. a. 25
  2. b. 20√3
  3. c. 30
  4. d. 25√3

Solution:

  1. Answer: (d)

    JEE Main Feb 24 Shift 1 2021 Solved Maths Papers

    tan θ = h/75 = 75/3h

    h2 = 752/3

    h = 25√3m


Question 12: If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

  1. a. –2t3
  2. b. –t3
  3. c. 0
  4. d. 2t3

Solution:

  1. Answer: (a)

    Equation of tangent at P(t, t3)

    (y - t3) = 3t2(x - t) ⋯(1)

    Now solve the above equation with

    y = x3 ⋯(2)

    By (1) & (2)

    x3 – t3 = 3t2 (x - t)

    x2 + xt + t2 = 3t2

    x2 + xt – 2t2 = 0

    ⇒(x – t)(x + 2t) = 0

    ⇒x = - 2t

    ⇒Q(-2t, -8t3)

    Ordinate of required point = (2t3 + (-8t)3)/3

    = -2t3


Question 13: The area (in sq.units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is:

  1. a. 24π + 3√3
  2. b. 12π + 3√3
  3. c. 12π - 3√3
  4. d. 24π - 3√3

Solution:

  1. Answer: (d)

    The curves intersect at point (3, ± 3√3)

    Shift 1 JEE Main Feb 24 2021 Solved Maths Papers

    Required area

    Shift 1 2021 JEE Main Feb 24 Solved Maths Papers


Question 14: If ∫(cos x -sin x)/√(8-sin 2x) dx = a sin-1(sin x + cos x)/b + c where c is a constant of integration, then the ordered pair (a, b) is equal to:

  1. a. (1, –3)
  2. b. (1, 3)
  3. c. (–1, 3)
  4. d. (3, 1)

Solution:

  1. Answer: (b)

    Put sin x + cos x = t ⇒1 + sin 2x = t2

    (cos x -sin x ) dx = dt

    ⇒I = ∫dt/√(8-(t2-1))

    = ∫dt/(9-t2)

    = sin-1 (t/3) + c = sin-1(sin x + cos x)/3 + c

    ⇒ a = 1 and b = 3


Question 15: The population P = P(t) at time ‘t’ of a certain species follows the differential equation dP/dt = 0.5P - 450. If P(0) = 850, then the time at which population becomes zero is:

  1. a. ½ loge 18
  2. b. 2loge 18
  3. c. loge 9
  4. d. loge 18

Solution:

  1. Answer: (b)

    dp/dt = (p-900)/2

    Shift 1 Feb 24 JEE Main 2021 Solved Maths Papers

    ln |900| - ln |50| = t/2

    t/2 = ln |18|

    ⇒ t = 2ln 18


Question 16: The value of - 15C1 + 2. 15C2 - 3.15C3+ ….-15.15C1 + 14C1+ 14C3 + 14C5 + ...14C11 is

  1. a. 214
  2. b. 213 - 13
  3. c. 216 - 1
  4. d. 213 - 14

Solution:

  1. Answer: (d)

    Maths Shift 1 JEE Main Feb 24 2021 Solved Papers

    = (14C1+ 14C3 + 14C5 + ...14C11 + 14C13) - 14C13

    = 213 - 14

    ∴ S1 + S2 = 213 - 14


Question 17: An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:

  1. a. 3/16
  2. b. 1/2
  3. c. 5/16
  4. d. 1/32

Solution:

  1. Answer: (b)

    P(odd no. twice) = P(even no. thrice)

    nC2 (1/2)n = nC3 (1/2)n

    ⇒ n = 5

    Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)

    = 5C1(1/2)5 + 5C3(1/2)5+ 5C5(1/2)5

    = 16/25

    = 1/2


Question 18: Let p and q be two positive number such that p + q = 2 and p4 + q4 = 272. Then p and q are roots of the equation:

  1. a. x2 - 2x + 2 = 0
  2. b. x2 - 2x + 8 = 0
  3. c. x2 - 2x + 136 = 0
  4. d. x2 - 2x + 16 = 0

Solution:

  1. Answer: (d)

    (p2 + q2)2 – 2p2q2 = 272

    ((p + q)2 - 2pq)2 - 2p2q2 = 272

    ⇒16 - 16pq + 2p2q2 = 272

    (pq)2 – 8pq –128 = 0

    ⇒pq = (8±24)/2 = 16, - 8

    ⇒pq = 16

    Now

    x2 - (p + q)x + pq = 0

    x2 - 2x + 16 = 0


Question 19: If e(cos2x+cos4x+cos6x+.....)loge2e^{(cos^{2}x + cos^{4}x + cos^{6}x + .....\infty )log_{e}2} satisfies the equation t2 - 9t + 8 = 0, then the value of 2sin x/(sin x + √3cos x) 0< x<π/2 is:

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

Solution:

  1. Answer: (c)

    Maths JEE Main Shift 1 Feb 24 2021 Solved Papers

    0 < x < π/2 ⇒ cot⁡x = √3

    ⇒(2 sin⁡x)/(sin⁡x+√3 cos⁡x )

    = 2/(1+√3 cot ⁡x)

    = 1/2


Question 20: The system of linear equations


3x – 2y – kz = 10

2x – 4y – 2z = 6

x + 2y – z = 5m

is inconsistent if :

  1. a. k = 3, m = 4/5
  2. b. k ≠ 3,m∈R
  3. c. k ≠ 3, m ≠ 4/5
  4. d. k = 3, m ≠ 4/5

Solution:

  1. Answer: (d)

    Maths JEE Main Feb 24 Shift 1 2021 Solved Papers

    ⇒ 3(4 + 4) + 2(–2 + 2) -k(4 + 4) = 0

    ⇒ k = 3

    Maths JEE Main Feb 24 Shift 1 Solved Paper 2021

    ⇒ 10(4 + 4) +2(-6 + 10m) -3(12 + 20m) ≠ 0

    ⇒ m ≠ 4/5

    Solved Paper Maths Shift 1 JEE Main Feb 24 2021

    ⇒ 3(-6 + 10m) - 10(- 2 + 2) - 3(10m - 6) ≠ 0

    ⇒ 0

    Solved Papers Maths Shift 1 Feb 24 2021 JEE Main

    ⇒3(-20m - 12) +2(10m - 6) +10(4 + 4) ≠ 0

    ⇒ m ≠ 4/5


Section B


Question 1: Let

Solved Papers Maths JEE Main Shift 1 Feb 24 2021

where α∈R. Suppose Q = [qij ] is a matrix satisfying PQ = kI3 for some non-zero k∈R. If q23= - k/8 and |Q⁡| = k2/2, then α2 + k2 is equal to ___

    Solution:

    1. Answer: 17

      Solved Papers Maths JEE Main Feb 24 Shift 1 2021


    Question 2: Let Bi (i = 1,2,3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α - 2β)p ⁡= αβ and (β - 3γ)p⁡= 2βγ (All the probabilities are assumed to lie in the interval (0, 1)). Then (P(B1⁡))/(P(B3⁡) )is equal to _____

      Solution:

      1. Answer: 6

        Let x,y,z be probability of B1, B2, B3 respectively.

        ⇒ x(1 – y) (1 – z) = α

        ⇒ y(1 – x)(1 – z) = β

        ⇒ z(1 – x)(1 – y ) = γ

        ⇒ (1 – x)(1 – y)(1 – z) = p

        Now (α - 2β)p = αβ

        ⇒ (x(1–y)(1–z)-2y(1-x)(1–z)) (1–x)(1–y)(1–z) = xy(1–x)(1–y) (1–z)2

        ⇒ x+ xy - 2y = xy

        ∴x = 2y⋯(1)

        Similarly, (β–3γ) p = 2βγ

        ⇒ y = 3z ⋯(2)

        From (1) & (2)

        ⇒x = 6z

        Hence x/z = P(B1)/P(B3) = 6


      Question 3: The minimum value of α for which the equation 4/sin⁡x +1/(1-sin⁡x )=α has at least one solution in (0,π/2) is _______

        Solution:

        1. Answer: 9

          f(x⁡) = 4/sin⁡x + 1/(1 - sin⁡x )

          Let sin⁡x = t ∵x∈(0, π/2) ⇒ 0 < t < 1

          f(t⁡)= 4/t + 1/(1-t)

          f'(t⁡) = (-4)/t2 + 1/(1 - t⁡)2 = 0

          ⇒(t2-4(1-t⁡)2/t2(1-t⁡)2 = 0

          ⇒t = 2/3

          fmin at t = 2/3

          αmin = f(2/3) =4/(2/3) + 1/(1 - 2/3)

          = 6 + 3

          = 9


        Question 4: If one of the diameters of the circle x2+y2 – 2x – 6y + 6 = 0 is a chord of another circle ‘C’ whose center is at (2,1), then its radius is ________

          Solution:

          1. Answer: 3

            Solved Papers Maths JEE Main 2021 Feb 24 Shift 1

            distance between (1,3) and (2,1) is √5

            ∴ (√5)2+(2)2= r2

            ⇒r = 3


          Question 5:


          Solution Papers Maths Shift 1 JEE Main Feb 24 2021

            Solution:

            1. Answer: 1

              Solution Papers Maths Shift 1 Feb 24 2021 JEE Main

              = tan 𝜋/4

              = 1


            Question 6: If aa(x+x2)dx=22\int_{-a}^{a}(\left | x \right |+\left | x-2 \right |)dx = 22 and [x] denotes the greatest integer ≤ x, then aa(x+[x])dx\int_{a}^{-a}(x+\left [ x \right ])dx is equal to

              Solution:

              1. Answer: 3

                Solution Papers Maths JEE Main Shift 1 Feb 24 2021


              Question 7: Let three vectors a\vec{a} ,b\vec{b} and c\vec{c} be such that c\vec{c} is coplanar with a\vec{a} and b\vec{b}, a.c=7\vec{a}.\vec{c}=7 and b\vec{b} is perpendicular to c\vec{c}, where a=i^+j^+k^\vec{a} = -\hat{i}+\hat{j}+\hat{k} and b=2i^+k^\vec{b} = 2\hat{i}+\hat{k}, then the value of 2a+b+c22\left | \vec{a} +\vec{b}+\vec{c} \right |^{2} is _______

                Solution:

                1. Answer: 75

                  Solution Papers Maths JEE Main Feb 24 Shift 1 2021


                Question 8: Let A = {n∈N∶ n is a 3-digit number}, B = {9k + 2∶ k∈N} and C = {9k + l∶ k∈N} for some l(0 < l < 9). If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to ___________

                  Solution:

                  1. Answer: 5

                    3 digit number of the form 9K+2 are {101,109,⋯,992}

                    ⇒ Sum equal to (100/2)(1093) = S1= 54650

                    Now 274 × 400 =S1+S2

                    ⇒274 × 400 = (100/2) [101 + 992] + S2

                    ⇒274 × 400 = 50 × 1093 + S2

                    ⇒S2 = 109600 – 54650

                    ∴S2 = 54950

                    S2 = 54950 = (100/2) [(99+l) + (990+l)]

                    ⇒ 2l + 1089 = 1099

                    ⇒ l = 5


                  Question 9: If the least and the largest real values of α, for which the equation z + α|z⁡-1| + 2 i⁡ = 0 (z∈C and i =√(-1)) has a solution, are p and q respectively; then 4(p2 + q2) is equal to _________

                    Solution:

                    1. Answer: 10

                      x + iy + α√((x–1)2 +y2) + 2i=0

                      ⇒y + 2 = 0 and x + α√((x-1)2+y2)=0

                      y = –2 & x2 = α2(x2 - 2x + 1 + 4)

                      α2 = x2/(x2- 2x + 5)

                      ⇒ x22 – 1) - 2xα2 + 5α2 = 0

                      ∵x∈R ⇒ D≥0

                      ⇒ 4α4 - 4(α2 - 1)5α2 ≥ 0

                      ⇒ α2 [4α2 - 20α2 + 20] ≥ 0

                      ⇒ α2 [-16α2 + 20] ≥ 0

                      ⇒ α22 - 5/4] ≤ 0

                      ⇒ α2 ∈ [0, 5/4]

                      ⇒ α ∈ [-√5/2, √5/2]

                      then 4[p2 + q2] = 4[5/4 + 5/4]

                      = 10


                    Question 10: Let M be any 3×3 matrix with entries from the set {0,1,2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is __________

                      Solution:

                      1. Answer: 540

                        Solution Papers Maths JEE Main 2021 Feb 24 Shift 1

                        ⇒ a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 = 7

                        Case I∶ Seven (1’s) and two (0’s)

                        9C2 = 36

                        Case II∶ One (2) and three (1’s) and five (0’s)

                        9!/5!3! = 504

                        ∴Total = 540


                      Video Solutions- February 2021 Question Papers

                      JEE Main 2021 Maths Paper February 24 Shift 1

                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 24 Shift 1