JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1

JEE Main 2021 Maths question paper for Feb 25 shift 1 has been added to this page with the main goal of helping students to properly understand the exact or overall nature of the JEE Main exam. The question paper can be directly viewed on this page or students also have the option to download the PDF version. After downloading, students are advised to start solving the questions and assess their preparation level. Moreover, after solving enough questions and getting enough practice students will develop better speed and precision to solve questions.
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JEE Main 2021 25th February Shift 1 Maths Question Paper

SECTION A


Question 1: The coefficients a,b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :

  1. a. 1/54
  2. b. 1/72
  3. c. 1/36
  4. d. 5/216

Solution:

  1. Answer: (d)

    ax2 + bx + c = 0

    a,b,c ∈ {1,2,3,4,5,6}

    n(s) = 6 × 6 × 6 = 216

    D=0 ⇒ b2 = 4ac

    ac = b2/4, If b = 2, ac = 1 ⇒ a = 1,c = 1

    If b = 4, ac = 4 ⇒ a = 1, c = 4

    a = 4, c = 1

    a = 2, c = 2

    If b = 6, ac = 9 ⇒ a = 3, c = 3

    Therefore, probability = 5/216


Question 2: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l2 + m2 – n2 = 0. Then the value of sin4 α + cos4 α is :

  1. a. 3/4
  2. b. 1/2
  3. c. 5/8
  4. d. 3/8

Solution:

  1. Answer: (c)

    Given that l+m=n ….(1)

    l2 + m2 – n2 = 0 ….(2)

    Squaring equation (1)

    l2 + m2 + 2lm = n2 ….(3)

    From equations (2) and (3)

    lm=0 ⇒ l = 0 or m = 0

    Case (1) : l = 0

    ⇒ m=n

    ⇒ l2 + m2 + n2 = 0

    ⇒ m=n = ±1/(√2 )

    ∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2)

    Case (2): m = 0

    ⇒ l = n

    ⇒ l2 + m2 + n2 = 0

    JEE MAIN 2021 Feb 25 Shift 1 Solution 2


Question 3: The value of the integral


JEE MAIN 2021 Feb 25 Shift 1 Solution 3

(where c is a constant of integration)

  1. a. 1/18[9 - 2 sin6 θ - 3 sin4 θ - 6 sin2 θ ](3/2) + c
  2. b. 1/18[11 - 18 sin2 θ + 9 sin4 θ - 2 sin6 θ ](3/2) + c
  3. c. 1/18[11 - 18 cos2 θ + 9 cos4 θ - 2 cos6 θ ](3/2) + c
  4. d. 1/18[9 - 2 cos6 θ - 3 cos4 θ – 6 cos2 θ ](3/2) + c

Solution:

  1. Answer: (c)

    Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin2A

    JEE MAIN 2021 Feb 25 Shift 1 Solved Question 3


Question 4: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

  1. a. 10(√3-1)
  2. b. 10√3
  3. c. 10
  4. d. 10(√3+1)

Solution:

  1. Answer: (d)

    JEE MAIN 2021 Feb 25 Shift 1 Solution 4

    x + y = √3h ….…….(1)

    Also,

    h/y = tan 45o

    h = y …….(2)

    put in (1)

    x + y = √3y

    x = (√3 – 1)y

    x/20=v (speed)

    Therefore, time taken to reach

    Foot from B

    = y/V

    = x/(√3-1)x . 20

    = 10 (√3 + 1)


Question 5:


JEE MAIN 2021 Feb 25 Shift 1 Solution 5

  1. a. xyz = 4
  2. b. xy – z = (x + y)z
  3. c. xy + yz + zx = z
  4. d. xy + z = (x+y)z

Solution:

  1. Answer: (d)

    JEE MAIN 2021 Feb 25 Shift 1 Solved Question 5


Question 6: The equation of the line through the point (0,1,2) and perpendicular to the line x12=y+13=z12\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2} is :

  1. a. x3=y14=z23\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}
  2. b. x3=y14=z23\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{3}
  3. c. x3=y14=z23\frac{x}{3} = \frac{y-1}{-4} = \frac{z-2}{3}
  4. d. x3=y14=z23\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{-3}

Solution:

  1. Answer: (a)

    x12=y+13=z12=λ\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}=\lambda

    Any point on this line (2λ + 1, 3λ - 1, -2λ + 1)

    Direction ratio of given line d1(2,3,2)\vec{d_{1}}\equiv (2, 3, -2)

    Direction ratio of the line to be found d2(2λ+1,3λ2,2λ1)\vec{d_{2}}\equiv (2\lambda + 1, 3\lambda -2, -2\lambda -1)

    ∴ d2.d2=0\vec{d_{2}}.\vec{d_{2}} = 0

    λ = 2 / 17

    Direction ratio of line (21, -28, -21) ≡ (3, -4, -3) ≡ (-3, 4, 3)


Question 7: The statement A→ (B → A) is equivalent to:

  1. a. A → (A ᐱ B)
  2. b. A → (A ᐯ B)
  3. c. A → (A → B) 
  4. d. A → (A ↔ B)

Solution:

  1. Answer: (b)

     A → (B → A) 

    ⇒ A → (∼ B ᐯ A)

    ⇒ ∼ A ᐯ (∼ B ᐯ A)

    ⇒ ∼ B ᐯ (∼ A ᐯ A)

    ⇒ ∼ B ᐯ t

    = t (tautology)

    From options:

    (b) A → (A ᐯ B)

    ⇒ ∼ A ᐯ (A ᐯ B)

    ⇒ (∼ A ᐯ A) ᐯ B

    ⇒ t ᐯ B

    ⇒ t


Question 8: The integer k, for which the inequality x2 – 2(3k – 1)x + 8k2 – 7 >0 is valid for every x in R is :

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

Solution:

  1. Answer: (a)

    D < 0

    (2(3k-1))2 – 4(8k2 – 7) < 0

    4(9k2 – 6k + 1) - 4(8k2 – 7) < 0

    k2 – 6 k + 8 < 0

    (k–4)(k–2)<0

    2< k < 4

    then k = 3


Question 9: A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?

  1. a. (0,3)
  2. b. (-6,0)
  3. c. (4,5)
  4. d. (5,4)

Solution:

  1. Answer: (d)

    Equation of tangent : y = mx + 3/(2m)

    mT = (1/2) (Because perpendicular to line 2x + y = 1)

    Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0

    Substitute given options in the above equation and check.

    We can find that (5, 4) does not lie on the line.

    Hence option (d) is the answer.


Question 10: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?

  1. a. f is one-one
  2. b. If fog is one-one, then g is one-one
  3. c. If g is onto, then fog is one-one
  4. d. If f is onto, then f(n) = n for all n ∈ N

Solution:

  1. Answer: (c)

    f(n + 1) = f(n) +f(1)

    ⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1)

    General term = Tn = f(1) + (n-1)f(1) = nf(1)

    ⇒f(n) = nf(1)

    Clearly f(n) is one-one.

    For fog to be one-one, g must be one-one.

    For f to be onto, f(n) should take all the values of natural numbers.

    As f(x) is increasing, f(1)=1

    ⇒f(n)=n

    If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.


Question 11: Let the lines (2 – i)z = (2 + i)zˉ\bar{z} and (2 + i)z + (i – 2)zˉ\bar{z} – 4i = 0, (here i2 = –1) be normal to a circle C. If the line iz + zˉ\bar{z} + 1 + i = 0 is tangent to this circle C, then its radius is :

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

Solution:

  1. Answer: (c)

    (2–i)z = (2+i) zˉ\bar{z}

    ⇒(2–i)(x+ iy)=(2+i)(x–iy)

    ⇒2x–ix + 2iy + y=2x+ ix–2iy+y

    ⇒2ix –4iy=0

    L1 ∶ x–2y=0

    ⇒(2+i)z+(i –2) zˉ\bar{z} –4i=0.

    ⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0.

    ⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0

    ⇒2ix + 4iy – 4i =0

    L2 ∶ x+2y–2 = 0

    Solve L1 and L2

    4y=2 or y=1/2

    ∴ x = 1

    Centre(1, 1/2)

    L3∶ iz + zˉ\bar{z} + 1 + i = 0

    ⇒i(x+iy)+x–iy+1+i=0

    ⇒ix–y+x–iy+1+i=0

    ⇒(x–y+1)+i(x–y+1)=0

    Radius = distance from (1, 1/2) to x – y + 1 = 0

    r = (1-1/2+1)/ √2

    r = 3/2√2


Question 12: All possible values of θ ∈ [0, 2π] for which sin2θ + tan 2θ > 0 lie in

  1. a. (0,π2)(π,3π2)\left ( 0,\frac{\pi }{2} \right )\cup \left ( \pi,\frac{3\pi }{2} \right )
  2. b. (0,π4)(π2,3π4)(π,5π4)(3π2,7π4)\left ( 0,\frac{\pi }{4} \right )\cup \left (\frac{\pi }{2},\frac{3\pi }{4} \right )\cup \left (\pi,\frac{5\pi }{4} \right )\cup \left (\frac{3\pi }{2},\frac{7\pi }{4} \right )
  3. c. (0,π2)(π2,3π4)(π,7π6)\left ( 0,\frac{\pi }{2} \right )\cup \left (\frac{\pi }{2},\frac{3\pi }{4} \right )\cup \left (\pi,\frac{7\pi }{6} \right )
  4. d. (0,π4)(π2,3π4)(3π2,11π6)\left ( 0,\frac{\pi }{4} \right )\cup \left (\frac{\pi }{2},\frac{3\pi }{4} \right )\cup \left (\frac{3\pi }{2},\frac{11\pi }{6} \right )

Solution:

  1. Answer: (b)

    JEE MAIN 2021 Feb 25 Shift 1 Solved Problem 12

    tan 2θ(1+cos 2θ) >0 (since 1+cos 2θ>0)

    2θ ∈ (0, π/2) ⋃(π, 3π/2) ⋃(2π, 5π/2)⋃(3π, 7π/2)

    => (0, π/4) ⋃(π/2, 3π/4) ⋃(π, 5π/4)⋃(3π/2, 7π/4)


Question 13: The image of the point (3,5) in the line x – y + 1 = 0, lies on :

  1. a. (x – 2)2 + (y – 4) 2 =4
  2. b. (x – 4) 2 + (y + 2)2 =16
  3. c. (x – 4)2 + (y – 4)2 = 8
  4. d. (x – 2)2 + (y – 2)2 =12

Solution:

  1. Answer: (a)

    Image of P(3, 5) on the line x – y + 1 = 0 is

    x31=y51=2(35+1)2=1\frac{x-3}{1} = \frac{y-5}{-1} = \frac{-2(3-5+1)}{2} = 1

    x = 4, y = 4

    Image is (4, 4)

    Which lies on (x – 2)2 + (y – 4)2 = 4


Question 14: If Rolle’s theorem holds for the function f(x) = x3 – ax2 + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :

  1. a. (–5, 8)
  2. b. (5, 8)
  3. c. (5, –8)
  4. d. (–5, –8)

Solution:

  1. Answer: (b)

    f(1) = f(2)

    ⇒ 1 – a + b –4 = 8 – 4a + 2b –4

    3a – b = 7 …(1)

    f’(x) = 3x2 – 2ax + b

    ⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0

    ⇒–8a + 3b = –16 …(2)

    a = 5, b = 8


Question 15: If the curves, x2a+y2b\frac{x^2}{a}+\frac{y^2}{b} and x2c+y2d\frac{x^2}{c}+\frac{y^2}{d} intersect each other at an angle of 90°, then which of the following relations is true ?

  1. a. a + b = c + d
  2. b. a- b = c – d
  3. c. ab = (c+d)/(a+b)
  4. d. a-c = b+d

Solution:

  1. Answer: (b)

    x2a+y2b\frac{x^2}{a}+\frac{y^2}{b} ………(1)

    diff : 2xa+2ybdydx=0\frac{2x}{a}+\frac{2y}{b}\frac{dy}{dx} = 0

    ybdydx=xa\frac{y}{b}\frac{dy}{dx} = \frac{-x}{a}

    dydx=bxay\frac{dy}{dx} = \frac{-bx}{ay} ……(2)

    x2c+y2d\frac{x^2}{c}+\frac{y^2}{d} ………(3)

    Diff : dydx=dxcy\frac{dy}{dx} = \frac{-dx}{cy} ………(4)

    m1m2 = –1 ⇒ bxay×dxcy=1\frac{-bx}{ay} \times \frac{-dx}{cy} = -1

    ⇒ bdx2 = – acy2…….(5)

    (1)–(3) ⇒ (1/a – 1/c)x2 + (1/b – 1/d) y2 = 0

    Solve above equation using 5

    ⇒ (c – a) – (d – b) = 0

    ⇒ c – a = d – b

    ⇒ c – d = a – b


Question 16:


JEE MAIN 2021 Feb 25 Shift 1 Solution 16

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

Solution:

  1. Answer: (c)

    JEE MAIN 2021 Feb 25 Shift 1 Solved Problem 16


Question 17: The total number of positive integral solutions (x, y, z) such that xyz = 24 is

  1. a. 36
  2. b. 45
  3. c. 24
  4. d. 30

Solution:

  1. Answer: (d)

    x.y.z = 24

    x.y.z=23.31

    x=2(a_1 )⋅3(b_1 )

    y = 2(a_2 )⋅3(b_2 )

    z = 2(a_3 )⋅3(b_3 )

    a1, a2, a3 ∈ {0,1,2,3}

    b1, b2, b3 ∈ {0,1}

    Case 1: a1 + a2 + a3 =3

    Non negative solution = (3+3-1) C(3-1) = 5C2 = 10

    Case 2: b1 + b2 + b3 = 1

    Non negative solution = (1+3-1) C(3-1) = 3C2 = 3

    ∴ Total solutions =10×3=30


Question 18: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is x24x+y+8x2\frac{x^2-4x+y+8}{x-2}, then this curve also passes through the point :

  1. a. (4, 5)
  2. b. (5, 4)
  3. c. (4, 4)
  4. d. (5, 5)

Solution:

  1. Answer: (d)

    dydx=(x2)2+y+4x2=(x2)+y+4x2\frac{dy}{dx} = \frac{(x-2)^2 +y+4}{x-2} = (x-2) + \frac{y+4}{x-2}

    Let x – 2 = t ⇒ dx = dt

    and y + 4 = u ⇒dy = du

    dy/dx = du/dt

    du/dt = t + u/t ⇒ du/dt - u/t = t

    Here, IF = 1/t

    u. (1/t) = ∫ t.(1\t) dt

    ⇒ u/t = t + c

    ⇒ (y+4)/(y-2) = (x – 2) + c

    Passing through (0, 0)

    c = 0

    ⇒ (y + 4) = (x – 2)2


Question 19: The value of 11x2ex3dx\int_{-1}^1 x^2 e^{x^3}dx where [t] denotes the greatest integer ≤ t,is :

  1. a. (e+1)/3
  2. b. (e-1)/3e
  3. c. (e+1)/3e
  4. d. 1/3e

Solution:

  1. Answer: (c)

    JEE MAIN 2021 Feb 25 Shift 1 Solution 19


Question 20: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

  1. a. 1/8
  2. b. 1/27
  3. c. 3/4
  4. d. 3/8

Solution:

  1. Answer:(a)

    Probability of not getting intercepted = 2/3

    Probability of missile hitting target = 3/4

    Probability that all 3 hit the target = (2/3 x3/4)3 = 1/8


Section B


Question 1: Let A1, A2, A3,.... be squares such that for each n ≥ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one is ____________.

    Solution:

    1. Answer: (9)

      JEE MAIN 2021 Feb 25 Shift 1 Solutions

      Side lengths are in GP.

      Tn = 12/(√2)n-1

      Area = 144/2n-1 <1

      => 2n-1 > 144

      Smallest n = 9


    Question 2: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to ___________

      Solution:

      1. Answer: (64)

        JEE MAIN 2021 Feb 25 Shift 1 Maths Solutions

        JEE MAIN 2021 Feb 25 Shift 1 Maths Solution


      Question 3: The locus of the point of intersection of the lines (√3)kx + ky - 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.

        Solution:

        1. Answer: (2)

          (√3)kx + ky - 4√3 = 0 ...(1)

          √3kx -ky= 4√3 k2 ...(2)

          Adding equation (1) & (2)

          2√3 kx = 4√3(k^2 + 1)

          x = 2 (k + 1/k) ....(3)

          Subtracting equation (1) & (2)

          y = 2√3(1/k - k) ………(4)

          x2/4 – y2/12 = 4

          x2/16 – y2/48 = 1 - Hyperbola

          e2 = 1 + 48/16

          or e = 2


        Question 4:


        JEE MAIN 2021 Feb 25 Shift 1 Maths Problems Solution

          Solution:

          1. Answer: 13

            Maths JEE MAIN 2021 Feb 25 Shift 1 Solutions

            |I2 + A|=|I2 - A|

            From equation (1)

            ∴a2 + b2 = 1

            ⇒13(a2 + b2) = 13


          Question 5: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = –1 and x = 1. If limx0f(x)x3=1\lim_{x \to 0}\frac{f(x)}{x^3}=1, then 5.f(2) is equal to _________.

            Solution:

            1. Answer: (144)

              f(x) = x6 + ax5 + bx4 + x3

              f’(x) = 6x5 + 5ax4 + 4bx3 + 3x2

              Roots 1 & – 1

              Therefore, 6 + 5a + 4b + 3 = 0 & – 6 + 5a – 4b + 3 = 0 solving

              a = -3/5 and b = -3/2

              f(x) = x6 – (3/5)x5 – (3/2)x4 + x3

              5.f(2) = 5 [64 – 96/5 – 24 + 8] = 144


            Question 6: The number of points, at which the function f(x) = |2x + 1| - 3|x+2|+|x2 + x–2|, x ∈ R is not differentiable, is ________.

              Solution:

              1. Answer: 2

                Maths JEE MAIN 2021 Feb 25 Shift 1 Problem Solution

                Check at 1, –2 and -1/2

                Non-Differentiable at x = 1 and -1/2


              Question 7: If the system of equations

              kx + y + 2z = 1

              3x – y – 2z = 2

              –2x – 2y – 4z = 3

              has infinitely many solutions, then k is equal to __________.

                Solution:

                1. Answer: 21

                  D = D1 = D2 = D3 = 0

                  Choose D2 = 0

                  k(–8+6)–1(–12–4)+2(9+4)=0

                  –2k + 16 + 26 = 0

                  2k = 42

                  k = 21


                Question 8: Let a=i^+2j^k^,b=i^j^\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} and c=i^j^k^\vec{c} = \hat{i} - \hat{j} - \hat{k} be three given vectors. If r\vec{r} is a vector such that r×a=c×a\vec{r} \times \vec{a} = \vec{c} \times \vec{a} and r.b=0\vec{r} . \vec{b} = 0, then r.a=0\vec{r} . \vec{a} = 0 is equal to _______

                  Solution:

                  1. Answer: 12

                    Maths 2021 JEE MAIN Feb 25 Shift 1 Solutions


                  Question 9: Let A = [xyzyzxzxy]\begin{bmatrix} x &y &z \\ y &z &x \\ z & x &y \end{bmatrix}, where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A2 = l3, then the value of x3 + y3 + z3 is ______.

                    Solution:

                    1. Answer: 7

                      Maths JEE MAIN 2021 Shift 1 Feb 25 Solutions


                    Question 10: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.

                      Solution:

                      1. Answer: 32

                        Maths JEE MAIN 2021 Shift 1 Feb 25 Paper Solutions


                      Video Solutions- February 2021 Question Papers

                      JEE Main 2021 Maths Paper February 25 Shift 1

                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1
                      JEE Main 2021 Maths Paper With Solutions Feb 25 Shift 1