JEE Main 2021 February 26 – Shift 2 Maths Question Paper with Solutions

JEE aspirants will find solutions for all the questions asked in JEE Main 2021 Mathematics paper for February 26th Shift 2 exam on this page. The set of solutions have been prepared by BYJU’s expert teachers. The solutions have been crafted in a detailed manner. Practising these questions will help every JEE aspirant to understand the overall exam pattern and prepare accordingly.

JEE Main 2021 26th February Shift 2 Maths Question Paper

SECTION A


Question 1: Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P (ɑ, β, γ) is the foot of the perpendicular from (3, 2, 1) on L, then the value of 21 (ɑ + β + γ) equals:

  1. a. 142
  2. b. 68
  3. c. 136
  4. d. 102

Solution:

  1. Answer: (d)

    Equation of the line is x + 2y + z - 6 = 0 = y + 2z = 4

    [x + 2] / 3 = [y - 4] / - 2 = z / 1 = λ

    Jee Main 2021 26 Feb Shift II Question 1

    Dr’s of PQ: 3λ - 5, - 2λ - 2, λ - 1

    Dr’s of y lines are (3, - 2, 1)

    Since PQ is perpendicular to the line.

    3 (3λ - 5) - 2 (- 2λ + 2) + 1 (λ - 1) = 0

    λ = 10 / 7

    P (16 / 7, 8 / 7, 10 / 7)

    21 (ɑ + β + γ) = 21 (34 / 7) = 102


Question 2: The sum of the series ∑n=1∞[n2 + 6n + 10] / (2n + 1)! is equal to:

  1. a. (41/ 8) e + (19 / 8) e-1 - 10
  2. b. (- 41/ 8) e + (19 / 8) e-1 - 10
  3. c. (41/ 8) e - (19 / 8) e-1 - 10
  4. d. (41/ 8) e + (19 / 8) e-1 + 10

Solution:

  1. Answer: (c)

    n=1∞[n2 + 6n + 10] / (2n + 1)!

    Put 2n + 1 = r, where r = 3, 5, 7…..

    Jee Main 2021 26 Feb Shift II Question 2

    n = [r - 1] / 2


Question 3: Let f(x) be a differentiable function at x = a with f’ (a) = 2 and f (a) = 4. Then limx→a [x f (a) - a f (x)] / [x - a] equals:

  1. a. 2a + 4
  2. b. 2a - 4
  3. c. 4 - 2a
  4. d. a + 4

Solution:

  1. Answer: (c)

    By L-H rule

    L = limx→a [f (a) - a f’ (x)] / [1]

    L = 4 - 2a


Question 4: Let A (1, 4) and B (1, - 5) be two points. Let P be a point on the circle (x - 1)2 + (y - 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points P, A and B lie on:

  1. a. a parabola
  2. b. a straight line
  3. c. a hyperbola
  4. d. an ellipse

Solution:

  1. Answer: (b)

    PA2 = cos2 θ + (sin θ - 3)2 = 10 - 6 sin θ

    PB2 = cos2 θ + (sin θ + 6)2 = 37 + 12 sin θ

    PA2 + PB2|max = 47 + 6 sin θ|max

    θ = π / 2

    The points P, A and B lie on the line x = 1.


Question 5: If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of the radius r, then r is equal to :

  1. a. 14
  2. b. 12
  3. c. 1
  4. d. 13

Solution:

  1. Answer: (b)

    P ≡ (2h - 3), (2k - 2) → on circle

    [h - (3 / 2)]2 + (k - 1)2 = 1 / 4

    Radius = 1 / 2


Question 6: Let the slope of the tangent line to a curve at any point P (x, y) be given by [xy2 + y] / x. If the curve intersects the line x + 2y = 4 at x = - 2, then the value of y, for which the point (3, y) lies on the curve, is:

  1. a. - 18 / 11
  2. b. - 18 / 19
  3. c. - 4 / 3
  4. d. 18 / 35

Solution:

  1. Answer: (b)

    dy / dx = [xy2 + y] / x

    [x dy - y dx] / y2 = x dx

    - d (x / y) = d (x2 / 2)

    (- x / y) = (x2 / 2) + c

    The curve intersects the line x + 2y = 4 at x = - 2.

    So, - 2 + 2y = 4

    y = 3

    The curve passes through (-2, 3).

    (2 / 3) = 2 + c

    c = - 4 / 3

    The curve is (- x / y) = (x2 / 2) - (4 / 3).

    It also passes through (3, y).

    (- 3 / y) = (9 / 2) - (4 / 3)

    y = - 18 / 19


Question 7: Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, the x-axis and x = π / 2 in the first quadrant. Then,

  1. a. A1 = A2 and A1 + A2 = √2
  2. b. A1 : A2 = 1 : 2 and A1 + A2 = 1
  3. c. 2A1 = A2 and A1 + A2 = 1 + √2
  4. d. A1 : A2 = 1 : √2 and A1 + A2 = 1

Solution:

  1. Answer: (d)

    A1 + A2 = ∫0π/2 cosx dx = sin x|0π/2 = 1

    Jee Main 2021 26 Feb Shift II Question 7

    A1 = ∫0π/4 (cosx - sinx) dx = (sinx + cosx)0π/4 = √2 - 1

    A2 = 1 - (√2 - 1) = 2 - √2

    A1 / A2 = (√2 - 1) / (√2 (√2 - 1)) = 1 / √2


Question 8: If 0 < a, b < 1, and tan-1 a + tan-1 b = π / 4, then the value of (a + b) - [a2 + b2] / 2 + [a3 + b3 ] / 3 - [a4 + b4] / 4 + …. is

  1. a. loge2
  2. b. loge (e / 2)
  3. c. e
  4. d. e2 - 1

Solution:

  1. Answer: (a)

    tan-1 {[a + b] / [1 - ab]} = (π / 4)

    a + b = 1 - ab

    (1 + a) (1 + b) = 2

    (a + b) - [a2 + b2] / 2 + [a3 + b3] / 3 +…infinity

    = [a - (a2 / 2) + (a3 / 3) …..] + [b - (b2 / 2) + (b3 / 3) …..]

    = loge (1 + a) + loge (1 + b)

    = loge (1 + a) (1 + b) = loge 2


Question 9: Let F1 (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F2 (A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then:

  1. a. F1 is not a tautology but F2 is a tautology
  2. b. F1 is a tautology but F2 is not a tautology
  3. c. F1 and F2 both area tautologies
  4. d. Both F1 and F2 are not tautologies

Solution:

  1. Answer: (a)

    F1 (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A

    Using the set theory

    (A ∩ B′) ∪ (C′ ∩ (A ∪ B)) ∪ A′ = (A − A ∩ B) ∪ (S − A) ∪ [(S − C) ∩ (A ∪ B)]

    = (S − A ∩ B) ∪ [A ∪ B − C ∩ (A ∪ B)]

    = S − A ∩ B ∩ C.

    Hence not a tautology.

    Now F2 (A, B) = (A ∨ B) ∨ (B → ~A) = (A ∨ B) ∨ (~B ∨ A)

    Using set theory (A ∪ B) ∪ (B′ ∪ A) = (A ∪ B) ∪ (S − A ∪ B) = S

    Hence it is a tautology.


Question 10: Consider the following system of equations:


x + 2y - 3z = a

2x + 6y - 11 z = b

x - 2y + 7z = c,

Where a, b and c are real constants. Then the system of equations:

  1. a. has a unique solution when 5a = 2b + c
  2. b. has an infinite number of solutions when 5a = 2b + c
  3. c. has no solution for all a, b and c
  4. d. has a unique solution for all a, b and c

Solution:

  1. Answer: (b)

    Jee Main 2021 26 Feb Shift II Question 10

    = 20a - 2(7b + 11c) - 3(- 2b - 6c)

    = 20a - 14b - 22c + 6b + 18c

    = 20a - 8b - 4c

    = 4(5a - 2b - c)

    Δ2=1a32b111c7\Delta _{2} = \begin{vmatrix} 1 & a & -3 \\ 2& b & -11\\ 1& c & 7 \end{vmatrix}

    = 7b+ 11c - a(25) - 3(2c - b)

    = 7b+ 11c - 25a - 6c + 3b

    = - 25a + 10b+ 5c

    = - 5(5a - 2b- c)

    Δ3=12a26b12c\Delta _{3} = \begin{vmatrix} 1 & 2 & a\\ 2& 6 & b\\ 1& -2 & c \end{vmatrix}

    = 6c + 2b- 2(2c - b) - 10a

    = - 10a + 4b+ 2c

    = - 2(5a - 2b- c)

    For infinite solution,

    Δ = Δ1 = Δ2 = Δ3 = 0

    ⇒ 5a = 2b + c


Question 11: A seven-digit number is formed using the digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is:

  1. a. 6 / 7
  2. b. 4 / 7
  3. c. 3 / 7
  4. d. 1 / 7

Solution:

  1. Answer: (3)

    n (S) = 7! / [2! 3! 2!]

    n (E) = 6! / [2! 2! 2!]

    P (E) = n (E) / n (S) = [6! / 7!] * [[2! 3! 2!] / [2! 2! 2!]] = 3 / 7


Question 12: If vectors a1 = xi - j + k and a2 = i + yj + zk are collinear, then a possible unit vector parallel to the vector xi + yj + zk is :

  1. a. (1 / √2) (- j + k)
  2. b. (1 / √2) (i - j)
  3. c. (1 / √3) (i - j + k)
  4. d. (1 / √3) (i + j - k)

Solution:

  1. Answer: (3)

    (x / 1) = - 1 / y = 1 / z = λ (let)

    Unit vector parallel to xi + yj + zk = ± [λi - (1 / λ) j + (1 / λ) k] / √λ2 + (2 / λ2)

    For λ = 1, it is ± [i - j + k] / √3


Question 13: For x > 0, if f (x) = ∫1x loge t / [1 + t] dt, then f (e) + f (1 / e) is equal to:

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

Solution:

  1. Answer: (a)

    Jee Main 2021 26 Feb Shift II Question 13


Question 14: Let f : R → R be defined as

Jee Main 2021 26 Feb Shift 2 Question 14

If f (x) is continuous on R, then a + b equals:

  1. a. 3
  2. b. -1
  3. c. -3
  4. d. 1

Solution:

  1. Answer: (b)

    If f is continuous at x = - 1, then f (- 1-) = f (- 1)

    2 = |a - 1 + b|

    |a + b - 1| = 2 --- (i)

    Similarly f (1-) = f (1)

    |a + b + 1| = 0

    a + b = - 1


Question 15: Let A = {1, 2, 3 ……. , 10} and f : A → A be defined as f(k)={k+1ifkisoddkifkiseven}f(k) = \begin{Bmatrix} k+1 & if\: k\: is\: odd\\ k & if\: k\: is\: even \end{Bmatrix}. Then the number of possible functions g : A → A such that gof = f is:

  1. a. 105
  2. b. 10C5
  3. c. 55
  4. d. 5!

Solution:

  1. Answer: (a)

    g (f (x)) = f (x)

    g (x) = x, when x is even.

    5 elements in A can be mapped to any 10

    So, 105 × 1 = 105


Question 16: A natural number has prime factorization given by n = 2x3y5z, where y and z are such that y + z = 5 and y-1 + z-1 = 5 / 6, y > z. Then the number of odd divisors of n, including 1, is:

  1. a. 11
  2. b. 6x
  3. c. 12
  4. d. 6

Solution:

  1. Answer: (c)

    y + z = 5 . . . (1)

    (1 / y) + (1 / z) = 5 / 6

    yz = 6

    Also, (y - z)2 = (y + z)2 - 4yz

    y - z = ± 1 --- (2)

    From (1) and (2), y = 3 or 2 and z = 2 or 3

    For calculating the odd divisor of p = 2x3y5z, x must be 0.

    P = 203352

    Total odd divisors must be (3 + 1) (2 + 1) = 12.


Question 17: Let f (x) = sin-1 x and g (x) = [x2 - x - 2] / [2x2 - x - 6]. If g (2) = limx→2 g (x), then then the domain of the function fog is:

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

Solution:

  1. Answer: (a)

    g (2) = limx→2 [(x - 2) (x + 1)] / [(2x + 3) (x - 2)] = 3 / 7

    For domain of f (g (x))

    |(x2 - x - 2) / (2x2 - x - 6)| ≤ 1 [because of f (x) is [- 1, 1])

    (x + 1)2 ≤ (2x +3)2

    3x2 + 10x + 8 ≥ 0

    (3x + 4) (x + 2) ≥ 0

    x belongs to (- ∞, - 2] ⋃ [- 4 / 3, ∞]


Question 18: If the mirror image of the point (1, 3, 5) with respect to the plane 4x - 5y + 2z = 8 is (ɑ, β, γ), then 5 (ɑ + β + γ) equals:

  1. a. 47
  2. b. 39
  3. c. 43
  4. d. 41

Solution:

  1. Answer: (a)

    Image of (1, 3, 5) in the plane 4x - 5y + 2z = 8 is (ɑ, β, γ).

    [ɑ - 1] / 4 = [β - 3] / - 5 = [γ - 5] / 2 = - 2 [4 (1) - 5 (3) + 2 (5) - 8] / [42 + 52 + 22] = 2 / 5

    ɑ = 1 + 4 (2 / 5) = 13 / 5

    β = 3 - 5 (2 / 5) = 1 = 5 / 5

    γ = 5 + 2 (2 / 5) = 29 / 5

    5 (ɑ + β + γ) = 5 [(13 / 5) + (5 / 5) + (29 / 5)] = 47


Question 19: Let f (x) = ∫0x et f (t) dt + ex be a differentiable function for all x ∈ R. Then f (x) equals

  1. a. 2eex-1 - 1
  2. b. e(e^x-1)
  3. c. 2ee^x - 1
  4. d. ee^x - 1

Solution:

  1. Answer: (a)

    Given, f (x) = ∫0x et f (t) dt + ex ---- (1)

    f’ (x) = ex . f (x) + ex [using Newton Leibnitz Theorem]

    [f’ (x)] / [f (x) + 1] = ex

    ln (f (x) + 1) = ex + c

    Put x = 0, ln 2 = 1 + c

    ln [f (x) + 1] = ex + ln 2 - 1

    f (x) + 1 = 2 eex-1

    f (x) = 2 eex-1 - 1


Question 20: The triangle of the maximum area that can be inscribed in a given circle of radius ‘r’ is:

  1. a. A right-angle triangle having two of its sides of length 2r and r.
  2. b. An equilateral triangle of height 2r / 3.
  3. c. isosceles triangle with base equal to 2r.
  4. d. An equilateral triangle having each of its side of length √3r.

Solution:

  1. Answer: (d)

    The triangle of the maximum area that can be inscribed in a circle is an equilateral triangle.

    Let △ABC be inscribed in the circle,

    Now, in △OBD

    OD = r cos 60o = r / 2

    Height = AD = 3r / 2

    Again in △ABD

    Now sin 60o = (3r / 2) / AB

    AB = √3r

    Jee Main 2021 26 Feb Shift II Question 20


Section - B


Question 1: The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _______

    Solution:

    1. Answer: 1000

      Since the required number has G.C.D with 18 as 3. It must be an odd multiple of ‘3’ but not a multiple of ‘9’.

      (i) Now, a 4-digit number which is an odd multiple of ‘3’ are,

      1005, 1011, 1017, ⋯ , 9999 → 1499

      (ii) The 4-digit number which is an odd multiple of 9 are,

      1017, 1035, ⋯ , 9999 → 499

      Required numbers = 1499 − 499 = 1000


    Question 2: Let ɑ and β be two real numbers such that ɑ + β = 1 and ɑ β = - 1. Let Pn = (ɑ)n + (β)n, Pn-1 = 11 and Pn+1 = 29 for some integer n≥ 1. Then, the value of Pn2 is______________.

      Solution:

      1. Answer: 324

        Given, ɑ + β = 1 and ɑ β = - 1

        A quadratic equation with roots ɑ, β is x2 - x - 1 = 0.

        ɑ2 = ɑ + 1

        Multiplying both sides by ɑn-1

        ɑn+1 = ɑn + ɑn-1 --- (1)

        Similarly,

        βn+1 = βn + βn-1 --- (2)

        Adding (1) and (2)

        ɑn+1 + βn+1 = (ɑ)n + (β)n + [ɑn-1 + βn-1]

        Pn+1 = Pn + Pn-1

        29 = Pn + 11

        Pn = 18

        Pn2 = 182 = 324


      Question 3: Let X1, X2, …. X18 be eighteen observation such that ∑i=118 (Xi - ɑ) = 36 and ∑i=118 (Xi - β)2 = 90, where ɑ and β are distinct real numbers. If the standard deviation of these observations is 1, then the value of |ɑ - β| is ___________.

        Solution:

        1. Answer: 4

          Given, ∑i=118 (Xi - ɑ) = 36

          ∑xi - 18ɑ = 36

          ∑xi = 18 (ɑ + 2) ---- (1)

          i=118 (Xi - β)2 = 90

          ∑xi2 + 18β2 - 2β ∑xi = 90

          ∑xi2 + 18β2 - 2β * 18 (ɑ + 2) = 90 (using equation (1))

          ∑xi2 = 90 - 18β2 + 36β (ɑ + 2)

          Now, 𝜎2 = 1 ⇒ (1 / 18) ∑xi2 - (∑xi / 18)2 = 1

          (1 / 18) [90 - 18β2 + 36ɑβ + 72β] - [18 (ɑ + 2) / 18]2 = 1

          5 − β2 + 2ɑβ + 4β − (ɑ + 2)2 = 1

          ⇒ 5 − β2 + 2ɑβ + 4β − ɑ2 − 4 − 4ɑ = 1

          ⇒ ɑ2 − β2 + 2ɑβ + 4β − 4ɑ = 0

          ⇒ (ɑ − β)(ɑ − β + 4) = 0

          ⇒ ɑ − β = −4

          Hence |ɑ − β| = 4 (ɑ ≠ β).


        Question 4: In Im,n = ∫01 xm-1 (1 - x)n-1 dx, for m, n ≥ 1 and ∫01 [xm-1 + xn-1] / [(1 + x)m+n] dx = ɑIm,n, ɑ belongs to R, then ɑ =

          Solution:

          1. Answer: 1

            Jee Main 2021 26 Feb Shift II Section B Question 4


          Question 5: Let L be a common tangent line to the curves 4x2 + 9y2 = 36 and (2x)2 + (2y)2 = 31. Then the square of the slope of the line L is ______________.

            Solution:

            1. Answer: 3

              E : (x2 / 9) + (y2 / 4) = 1

              C : x2 + y2 = 31 / 4

              The equation of the tangent to the ellipse is y = mx ± √(9m2 + 4) --- (1)

              The equation of the tangent to the circle is y = mx ± √[(31 / 4) m2 + (31 / 4)] --- (2)

              Comparing equation (1) & (2), 9m2 + 4 = (31 / 4) m2+ (31 / 4)

              36m2 + 16 = 31m2 + 31

              5m2 = 15

              m2 = 3


            Question 6: If the matrix A=[100020301]A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 3 & 0 & -1 \end{bmatrix} satisfies the equation A20 + ɑA19 + βA = [100040001]\begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 1 \end{bmatrix} for some real numbers ɑ and β, then β -ɑ is equal to_________.

              Solution:

              1. Answer: 4

                Jee Main 2021 26 Feb Shift II Section B Question 6 solution


              Question 7: If the arithmetic mean and the geometric mean of the pth and qth terms of the sequence - 16, 8, - 4, 2, … satisfy the equation 4x2 - 9x + 5 = 0, then p + q is equal to ______________.

                Solution:

                1. Answer: 10

                  Given, 4x2 - 9x + 5 = 0

                  (x - 1) (4x - 5) = 0

                  AM = 5 / 4, GM = 1 (AM ≥ GM)

                  Again for the series, - 16, 8, - 4, 2, …

                  pth term tp = - 16 (- 1 / 2)p-1

                  qth term tq = - 16 (- 1 / 2)q-1

                  AM = [tp + tq] / 2 = 5 / 4

                  GM = √(tptq) = 1

                  162 (- 1 / 2)p+q-2 = 1

                  (- 2)8 = (- 2)p+q-2

                  p + q = 10


                Question 8: Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, - 3) and (4, - 2 √2), and given that a - 2√2b = 3, then (a2 + b2 + ab) is equal to______________.

                  Solution:

                  1. Answer: 9

                    Let the equation of normal is Y - y = (- 1 / m) (X - x), where m = dy / dx

                    As it passes through (a, b)

                    b - y = (- 1 / m) (a - x) = - dx / dy [a - x]

                    (b - y) dy = (x - a) dx

                    by - (y2 / 2) = (x2 / 2) - ax + c --- (i)

                    It passes through (3, - 3) & (4, - 2 √2)

                    -3b - (9 / 2) = (9 / 2) - 3a + c

                    3a - 3b - c = 9 --- (ii)

                    Also - 2√2b - 4 = 8 - 4a + c

                    4a - 2 √2b - c = 12 --- (iii)

                    Also a - 2 √2b = 3 --- (iv) (given)

                    From (ii) - (iii), - a + (2√2 - 3) b = - 3 ---- (v)

                    From (iv) + (v), b = 0, a = 3

                    a2 + b2 + ab = 9


                  Question 9: Let z be those complex number which satisfies |z + 5| ≤ 4 and z (1 + i) + zˉ\bar{z} (1 - i) ≥ - 10, i = √-1. If the maximum value of |z + 1|2 is ɑ + β√2, then the value of ɑ + β is ________.

                    Solution:

                    1. Answer: 48

                      Given, |z + 5| ≤ 4

                      (x + 5)2 + y2 ≤ 16

                      Also, z (1 + i) + zˉ\bar{z} (1 - i) ≥ - 10

                      x - y ≥ - 5 ---- (2)

                      From (1) and (2), the locus of z is the shaded region in the diagram.

                      Jee Main 2021 26 Feb Shift II Section B Question 9

                      |z + 1| represents the distance of ‘z’ from Q (- 1, 0).

                      Clearly, p is the required position of z when |z + 1| is maximum.

                      P ≡ (- 5 - 2 √2, - 2 √2)

                      PQ2 = 32 + 16 √2

                      ɑ = 32

                      β = 16

                      ɑ + β = 48


                    Question 10: Let a be an integer such that all the real roots of the polynomial 2x5 + 5x4 + 10x3 + 10x2 + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to _______.

                      Solution:

                      1. Answer: 2

                        Let f (x) = 2x5 + 5x4 + 10x3 + 10x2 + 10x + 10

                        f’ (x) = 10 (x4 + 2x3 + 3x2 + 2x + 1)

                        = 10x2 [x2 + (1 / x2) + 2 [x + (1 / x)] + 3]

                        = 10x2 [(x + (1/x))2 + 2 (x + (1 / x)) + 1]

                        = 10x2 [[x + (1 / x)] + 1]2 > 0, for all x belongs to R

                        f (x) is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

                        By observation, f (- 1) = 3 > 0

                        f (- 2) = - 64 + 80 - 80 + 40 - 20 + 10

                        = - 34 < 0

                        f (x) has at least one root in (- 2, - 1) = (a, a + 1)

                        |a| = 2


                      Video Solutions- February 2021 Question Papers

                      JEE Main 2021 Maths Paper February 26 Shift 2

                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2
                      JEE Main 2021 Maths Paper With Solutions Feb 26 Shift 2